Manifolds, tensor analysis, and applications

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Manifolds, Tensor Analysis, and Applications Third Edition Jerrold E. Marsden

Tudor Ratiu

Control and Dynamical Systems 107–81

D´epartement de Math´ematiques

California Institute of Technology

´ Ecole polytechnique federale de Lausanne

Pasadena, California 91125

CH - 1015 Lausanne, Switzerland

with the collaboration of Ralph Abraham Department of Mathematics University of California, Santa Cruz Santa Cruz, California 95064

7 March 2007

ii

Library of Congress Cataloging in Publication Data Marsden, Jerrold Manifolds, tensor analysis and applications, Third Edition (Applied Mathematical Sciences) Bibliography: p. 631 Includes index. 1. Global analysis (Mathematics) 2. Manifolds(Mathematics) 3. Calculus of tensors. I. Marsden, Jerrold E. II. Ratiu, Tudor S. III. Title. IV. Series. QA614.A28 1983514.382-1737 ISBN 0-201-10168-S American Mathematics Society (MOS) Subject Classification (2000): 34, 37, 58, 70, 76, 93 Copyright 2001 by Springer-Verlag Publishing Company, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher, Springer-Verlag Publishing Company, Inc., 175 Fifth Avenue, New York, N.Y. 10010.

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Contents

Preface 1 Topology 1.1 Topological Spaces . Metric Spaces . . . . 1.2 1.3 Continuity . . . . . . 1.4 Subspaces, Products, Compactness . . . . 1.5 1.6 Connectedness . . . 1.7 Baire Spaces . . . . .

1

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3 3 10 14 17 22 28 33

2 Banach Spaces and Differential Calculus Banach Spaces . . . . . . . . . . . . . . . . . 2.1 2.2 Linear and Multilinear Mappings . . . . . . . The Derivative . . . . . . . . . . . . . . . . . 2.3 2.4 Properties of the Derivative . . . . . . . . . . 2.5 The Inverse and Implicit Function Theorems

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37 37 51 68 74 102

3 Manifolds and Vector Bundles 3.1 Manifolds . . . . . . . . . . . . . . . . . . . . 3.2 Submanifolds, Products, and Mappings . . . . 3.3 The Tangent Bundle . . . . . . . . . . . . . . 3.4 Vector Bundles . . . . . . . . . . . . . . . . . 3.5 Submersions, Immersions, and Transversality The Sard and Smale Theorems . . . . . . . . 3.6

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127 127 136 142 151 175 195

. . . . . . . . . . . . . . . . . . . . . . . . . . . and Quotients . . . . . . . . . . . . . . . . . . . . . . . . . . .

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4 Vector Fields and Dynamical Systems 211 4.1 Vector Fields and Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 4.2 Vector Fields as Differential Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234 4.3 An Introduction to Dynamical Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261

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Contents

4.4

Frobenius’ Theorem and Foliations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284

5 An Introduction to Lie Groups 295 5.1 Basic Definitions and Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295 Some Classical Lie Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 5.2 5.3 Actions of Lie Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 328 6 Tensors 6.1 Tensors on Linear Spaces . . . . . . . . Tensor Bundles and Tensor Fields . . . . 6.2 6.3 The Lie Derivative: Algebraic Approach The Lie Derivative: Dynamic Approach . 6.4 6.5 Partitions of Unity . . . . . . . . . . . .

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341 341 350 358 367 372

7 Differential Forms Exterior Algebra . . . . . . . . . . . . . . . . . . . . . . . . 7.1 7.2 Determinants, Volumes, and the Hodge Star Operator . . . 7.3 Differential Forms . . . . . . . . . . . . . . . . . . . . . . . . 7.4 The Exterior Derivative, Interior Product, & Lie Derivative 7.5 Orientation, Volume Elements and the Codifferential . . . .

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385 385 393 405 410 434

8 Integration on Manifolds 8.1 The Definition of the Integral . . . . . . . . . . . . . Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . 8.2 8.3 The Classical Theorems of Green, Gauss, and Stokes 8.4 Induced Flows on Function Spaces and Ergodicity . 8.5 Introduction to Hodge–deRham Theory . . . . . . .

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447 447 458 482 490 509

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529 529 548 560 568 580

9 Applications 9.1 Hamiltonian Mechanics . 9.2 Fluid Mechanics . . . . . 9.3 Electromagnetism . . . . The Lie–Poisson Bracket 9.4 9.5 Constraints and Control

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Contents

Page 1

Preface

This version: January 4, 2007

The purpose of this book is to provide core material in nonlinear analysis for mathematicians, physicists, engineers, and mathematical biologists. The main goal is to provide a working knowledge of manifolds, dynamical systems, tensors and differential forms. Some applications to Hamiltonian mechanics, fluid mechanics, electromagnetism, plasma dynamics and control theory are given in Chapter 9, using both invariant and index notation. Throughout the text there are clearly identified supplementary topics that may be omitted on a first reading or by readers who do not need to delve so deeply into certain topics. This device enables the reader to skip various topics without disturbing the main flow of the text. Some of these provide additional background material intended for completeness, to minimize the necessity of consulting too many outside references. Philosophy. We treat finite and infinite-dimensional manifolds simultaneously. This is partly for efficiency of exposition. Without advanced applications, using manifolds of mappings (such as applications to fluid dynamics), the study of infinite-dimensional manifolds can be hard to motivate. Chapter 9 gives an introduction to these applications. Some readers may wish to skip the infinite-dimensional case altogether. To aid in this, we have separated some of the technical points peculiar to the infinite-dimensional case into supplements, either directly in the text or on-line. Our own research interests lean toward physical applications, and the choice of topics is partly shaped by what has been useful to us over the years. We have tried to be as sympathetic to our readers as possible by providing ample examples, exercises, and applications. When a computation in coordinates is easiest, we give it and do not hide things behind complicated invariant notation. On the other hand, index-free notation sometimes provides valuable geometric and computational insight so we have tried to simultaneously convey this flavor. Prerequisites and Links. The prerequisites required are solid undergraduate courses in linear algebra and advanced calculus along with the usual mathematical maturity. At various points in the text contacts are made with other subjects. This provides a good way for students to link this material with other courses. For example, Chapter 1 links with point-set topology, parts of Chapters 2 and 8 are connected with functional analysis, Section 4.3 relates to ordinary differential equations and dynamical systems, Chapter 3 and Section 8.5 are linked to differential topology and algebraic topology, and Chapter 9 on applications is connected with applied mathematics, physics, and engineering.

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Preface

Use in Courses. This book is intended to be used in courses as well as for reference. The sections are, as far as possible, lesson sized, if the supplementary material is omitted. For some sections, like 2.5, 4.2, or 8.5, two lecture hours are required if they are to be taught in detail. A standard course for mathematics graduate students could omit Chapter 1 and the supplements entirely and do Chapters 2 through 8 in one semester with the possible exception of Section 8.4. The instructor could then assign certain supplements for reading and choose among the applications of Chapter 9 according to taste. A shorter course, or a course for advanced undergraduates, probably should omit all supplements, spend about two lectures on Chapter 1 for reviewing background point set topology, and cover Chapters 2 through 8 with the exception of Sections 4.4, 8.4, 8.5 and all the material relevant to volume elements induced by metrics, the Hodge star, and codifferential operators in Sections 6.2, 6.4, 6.5, and 8.2. A more applications oriented course could skim Chapter 1, review without proofs the material of Chapter 2 and cover Chapters 3 to 9 omitting the supplementary material and Sections 8.4 and 8.5. For such a course the instructor should keep in mind that while Sections 9.1 and 9.2 use only elementary material, Section 9.3 relies heavily on the Hodge star and codifferential operators, and Section 9.4 consists primarily of applications of Frobenius’ theorem dealt with in Section 4.4. The notation in the book is as standard as conflicting usages in the literature allow. We have had to compromise among utility, clarity, clumsiness, and absolute precision. Some possible notations would have required too much interpretation on the part of the novice while others, while precise, would have been so dressed up in symbolic decorations that even an expert in the field would not recognize them. History and Credits. In a subject as developed and extensive as this one, an accurate history and crediting of theorems is a monumental task, especially when so many results are folklore and reside in private notes. We have indicated some of the important credits where we know of them, but we did not undertake this task systematically. We hope our readers will inform us of these and other shortcomings of the book so that, if necessary, corrected printings will be possible. The reference list at the back of the book is confined to works actually cited in the text. These works are cited by author and year like this: deRham [1955]. Acknowledgements. During the preparation of the book, valuable advice was provided by Malcolm Adams, Morris Hirsch, Sameer Jalnapurkar, Jeff Mess, Charles Pugh, Clancy Rowley, Alan Weinstein, and graduate students in mathematics, physics and engineering at Berkeley, Santa Cruz, Caltech and Lausanne. Our other teachers and collaborators from whom we learned the material and who inspired, directly and indirectly, various portions of the text are too numerous to mention individually, so we hereby thank them all collectively. We have taken the opportunity in this edition to correct some errors kindly pointed out by our readers and to rewrite numerous sections. We thank Connie Calica, Dotty Hollinger, Anne Kao, Marnie MacElhiny and Esther Zack for their excellent typesetting of the book. We also thank Hendra Adiwidjaja, Nawoyuki Gregory Kubota, Robert Kochwalter and Wendy McKay for the typesetting and figures for this third edition. Jerrold E. Marsden and Tudor S. Ratiu January, 2007

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1 Topology

This version: January 4, 2007

The purpose of this chapter is to introduce just enough topology for later requirements. It is assumed that the reader has had a course in advanced calculus and so is acquainted with open, closed, compact, and connected sets in Euclidean space (see for example Marsden and Hoffman [1993]). If this background is weak, the reader may find the pace of this chapter too fast. If the background is under control, the chapter should serve to collect, review, and solidify concepts in a more general context. Readers already familiar with point set topology can safely skip this chapter. A key concept in manifold theory is that of a differentiable map between manifolds. However, manifolds are also topological spaces and differentiable maps are continuous. Topology is the study of continuity in a general context, so it is appropriate to begin with it. Topology often involves interesting excursions into pathological spaces and exotic theorems that can consume lifetimes. Such excursions are deliberately minimized here. The examples will be ones most relevant to later developments, and the main thrust will be to obtain a working knowledge of continuity, connectedness, and compactness. We shall take for granted the usual logical structure of analysis, including properties of the real line and Euclidean space

1.1

Topological Spaces

The notion of a topological space is an abstraction of ideas about open sets in Rn that are learned in advanced calculus. 1.1.1 Definition. sets such that

A topological space is a set S together with a collection O of subsets of S called open

T1. ∅ ∈ O and S ∈ O; T2. if U1 , U2 ∈ O, then U1 ∩ U2 ∈ O; T3. the union of any collection of open sets is open. The Real Line and n-space. For the real line with its standard topology, we choose S = R, with O, by definition, consisting of all sets that are unions of open intervals. Here is how to prove that this is a

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1. Topology

topology. As exceptional cases, the empty set ∅ ∈ O and R itself belong to O. Thus, T1 holds. For T2, let U1 and U2 ∈ O; to show that U1 ∩ U2 ∈ O, we can suppose that U1 ∩ U2 6= ∅. If x ∈ U1 ∩ U2 , then x lies in an open interval ]a1 , b1 [ ⊂ U1 and also in an interval ]a2 , b2 [ ⊂ U2 . We can write ]a1 , b1 [ ∩ ]a2 , b2 [ = ]a, b[ where a = max(a1 , a2 ) and b = min(b1 , b2 ). Thus x ∈ ]a, b[ ⊂ U1 ∩ U2 . Hence U1 ∩ U2 is the union of such intervals, so is open. Finally, T3 is clear by definition. Similarly, Rn may be topologized by declaring a set to be open if it is a union of open rectangles. An argument similar to the one just given for R shows that this is a topology, called the standard topology on Rn . The Trivial and Discrete Topologies. The trivial topology on a set S consists of O = {∅, S}. The discrete topology on S is defined by O = { A | A ⊂ S }; that is, O consists of all subsets of S. Closed Sets. Topological spaces are specified by a pair (S, O); we shall, however, simply write S if there is no danger of confusion. 1.1.2 Definition. Let S be a topological space. A set A ⊂ S will be called closed if its complement S\A is open. The collection of closed sets is denoted C. For example, the closed interval [0, 1] ⊂ R is closed because it is the complement of the open set ]−∞, 0[ ∪ ]1, ∞[. 1.1.3 Proposition.

The closed sets in a topological space S satisfy:

C1. ∅ ∈ C and S ∈ C; C2. if A1 , A2 ∈ C then A1 ∪ A2 ∈ C; C3. the intersection of any collection of closed sets is closed. Proof.

Condition C1 follows from T1 since ∅ = S\S and S = S\∅. The relations ! S\(A1 ∪ A2 ) = (S\A1 ) ∩ (S\A2 )

and S\

\

Bi

i∈I

=

[

(S\Bi )

i∈I

for {Bi }i∈I a family of closed sets show that C2 and C3 are equivalent to T2 and T3, respectively.



Closed rectangles in Rn are closed sets, as are closed balls, one-point sets, and spheres. Not every set is either open or closed. For example, the interval [0, 1[ is neither an open nor a closed set. In the discrete topology on S, any set A ⊂ S is both open and closed, whereas in the trivial topology any A 6= ∅ or S is neither. Closed sets can be used to introduce a topology just as well as open ones. Thus, if C is a collection satisfying C1–C3 and O consists of the complements of sets in C, then O satisfies T1–T3. Neighborhoods.

The idea of neighborhoods is to localize the topology.

1.1.4 Definition. An open neighborhood of a point u in a topological space S is an open set U such that u ∈ U . Similarly, for a subset A of S, U is an open neighborhood of A if U is open and A ⊂ U . A neighborhood of a point (or a subset) is a set containing some open neighborhood of the point (or subset). Examples of neighborhoods of x ∈ R are ]x − 1, x + 3], ]x − , x + [ for any  > 0, and R itself; only the last two are open neighborhoods. The set [x, x + 2[ contains the point x but is not one of its neighborhoods. In the trivial topology on a set S, there is only one neighborhood of any point, namely S itself. In the discrete topology any subset containing p is a neighborhood of the point p ∈ S, since {p} is an open set.

1.1 Topological Spaces

5

First and Second Countable Spaces. 1.1.5 Definition. A topological space is called first countable if for each u ∈ S there is a sequence {U1 , U2 , . . . } = {Un } of neighborhoods of u such that for any neighborhood U of u, there is an integer n such that Un ⊂ U . A subset B of O is called a basis for the topology, if each open set is a union of elements in B. The topology is called second countable if it has a countable basis. Most topological spaces of interest to us will be second countable. For example Rn is second countable since it has the countable basis formed by rectangles with rational side length and centered at points all of whose coordinates are rational numbers. Clearly every second-countable space is also first countable, but the converse is false. For example if S is an infinite non-countable set, the discrete topology is not second countable, but S is first countable, since {p} is a neighborhood of p ∈ S. The trivial topology on S is second countable (see Exercises 1.1-9 and 1.1-10 for more interesting counter-examples). 1.1.6 Lemma (Lindel¨ of’s Lemma). Every covering of a set A in a second countable space S by a family of open sets Ua (i.e., ∪a Ua ⊃ A) contains a countable subcollection also covering A. Proof. Let B = {Bn } be a countable basis for the topology of S. For each p ∈ A there are indices n and α such that p ∈ Bn ⊂ Uα . Let B 0 = { Bn | there exists an α such that Bn ⊂ Uα }. Now let Uα(n) be one of the Uα that includes the element Bn of B 0 . Since B 0 is a covering of A, the countable collection {Uα(n) } covers A.  Closure, Interior, and Boundary. 1.1.7 Definition. Let S be a topological space and A ⊂ S. The closure of A, denoted cl(A) is the intersection of all closed sets containing A. The interior of A, denoted int(A) is the union of all open sets contained in A. The boundary of A, denoted bd(A) is defined by bd(A) = cl(A) ∩ cl(S\A). By C3, cl(A) is closed and by T3, int(A) is open. Note that as bd(A) is the intersection of closed sets, bd(A) is closed, and bd(A) = bd(S\A). On R, for example, cl([0, 1[) = [0, 1],

int([0, 1[) = ]0, 1[,

and

bd([0, 1[) = {0, 1}.

The reader is assumed to be familiar with examples of this type from advanced calculus. 1.1.8 Definition. A subset A of S is called dense in S if cl(A) = S, and is called nowhere dense if S\ cl(A) is dense in S. The space S is called separable if it has a countable dense subset. A point u in S is called an accumulation point of the set A if each neighborhood of u contains a point of A other than itself. The set of accumulation points of A is called the derived set of A and is denoted by der(A). A point of A is said to be isolated if it has a neighborhood in A containing no other points of A than itself. The set A = [0, 1[ ∪ {2} in R has the element 2 as its only isolated point, its interior is int(A) = ]0, 1[, cl(A) = [0, 1] ∪ {2}, and der(A) = [0, 1]. In the discrete topology on a set S, int{p} = cl{p} = {p}, for any p ∈ S. Since the set Q of rational numbers is dense in R and is countable, R is separable. Similarly Rn is separable. A set S with the trivial topology is separable since cl{p} = S for any p ∈ S. But S = R with the discrete topology is not separable since cl(A) = A for any A ⊂ S. Any second-countable space is separable, but the converse is false; see Exercises 1.1-9 and 1.1-10. 1.1.9 Proposition.

Let S be a topological space and A ⊂ S. Then

(i) u ∈ cl(A) iff for every neighborhood U of u, U ∩ A 6= ∅; (ii) u ∈ int(A) iff there is a neighborhood U of u such that U ⊂ A;

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1. Topology

(iii) u ∈ bd(A) iff for every neighborhood U of u, U ∩ A 6= ∅ and U ∩ (S\A) 6= ∅. Proof. (i) u 6∈ cl(A) iff there exists a closed set C ⊃ A such that u 6∈ C. But this is equivalent to the existence of a neighborhood of u not intersecting A, namely S\C. (ii) and (iii) are proved in a similar way.  1.1.10 Proposition.

Let A, B and Ai , i ∈ I be subsets of S. Then

(i) A ⊂ B implies int(A) ⊂ int(B), cl(A) ⊂ cl(B), and der(A) ⊂ der(B); (ii) S\ cl(A) = int(S\A), S\ int(A) = cl(S\A), and cl(A) = A ∪ der(A) = A ∪ bd(A); (iii) cl(∅) = int(∅) = ∅, cl(S) = int(S) = S, cl(cl(A)) = cl(A), and int(int(A)) = int(A); (iv) cl(A ∪ B) = cl(A) ∪ cl(B), der(A ∪ B) = der(A) ∪ der(B), and int(A ∪ B) ⊃ int(A) ∪ int(B); (v) cl(A ∩ B) ⊂ cl(A) ∩ cl(B), der(A ∩ B) ⊂ der(A) ∩ der(B), and int(A ∩ B) = int(A) ∩ int(B); S S T T (vi) cl( Si∈I Ai) ⊃ Si∈I cl(Ai ), cl( i∈I Ai )T⊂ i∈I cl(A Ti ), int( i∈I Ai ) ⊃ i∈I int(Ai ), and int( i∈I Ai ) ⊂ i∈I int(Ai ). Proof. of the definition and of Proposition 1.1.9. Since for each S (i), (ii), and (iii) are consequences S S S T i ∈ I, Ai ⊂ i∈I Ai , by (i) cl(Ai ) ⊂ cl( i∈I Ai )Tand hence i∈I cl(Ai ) ⊂ cl( i∈I Ai ).TSimilarly, since i∈I Ai ⊂ Ai ⊂ cl(Ai ) for each i ∈ I, it follows that i∈I (Ai ) is a subset of the closed set i∈I cl(Ai ); thus by (i) ! ! \ \ \ cl Ai ⊂ cl cl(Ai ) = (cl(Ai )) . i∈I

i∈I

i∈I

The other formulas of (vi) follow from these and (ii). This also proves all the other formulas in (iv) and (v) except the ones with equalities. Since cl(A) ∪ cl(B) is closed by C2 and A ∪ B ⊂ cl(A) ∪ cl(B), it follows by (i) that cl(A ∪ B) ⊂ cl(A) ∪ cl(B) and hence equality by (vi). The formula int(A ∩ B) = int(A) ∩ int(B) is a corollary of the previous formula via (ii).  The inclusions in the above proposition can be strict. For example, if we let A = ]0, 1[ and B = [1, 2[ , then one finds that cl(A) = der(A) = [0, 1], cl(B) = der(B) = [1, 2], int(A) = ]0, 1[, int(B) = ]1, 2[, A ∪ B = ]0, 2[, and A ∩ B = ∅, and therefore int(A) ∪ int(B) = ]0, 1[ ∪ ]1, 2[ 6= ]0, 2[ = int(A ∪ B), and cl(A ∩ B) = ∅ 6= {1} = cl(A) ∩ cl(B). Let An = ]−1/n, 1/n[, n = 1, 2, . . . . Then \

An = {0},

int(An ) = An

n≥1

for all n, and int

\

 An

= ∅ 6= {0} =

n≥1

\

int(An ).

n≥1

Dualizing this via (ii) gives [ n≥1

cl(R\An ) = R\{0} 6= R = cl

[

 (R\An ) .

n≥1

If A ⊂ B, there is, in general, no relation between the sets bd(A) and bd(B). For example, if A = [0, 1] and B = [0, 2], A ⊂ B, yet we have bd(A) = {0, 1} and bd(B) = {0, 2}.

1.1 Topological Spaces

Convergence and Limit Points. straightforward way.

7

The notion of a convergent sequence carries over from calculus in a

1.1.11 Definition. Let S be a topological space and {un } a sequence of points in S. The sequence is said to converge if there is a point u ∈ S such that for every neighborhood U of u, there is an N such that n ≥ N implies un ∈ U . We say that un converges to u, or u is a limit point of {un }. For example, the sequence {1/n} ∈ R converges to 0. It is obvious that limit points of sequences un of distinct points are accumulation points of the set {un }. In a first countable topological space any accumulation point of a set A is a limit of a sequence of elements of A. Indeed, if {Un } denotes the countable collection of neighborhoods of a ∈ der(A) given by Definition 1.1.5, then choosing for each n an element an ∈ Un ∩ A such that an 6= a, we see that {an } converges to a. We have proved the following. 1.1.12 Proposition. Let S be a first-countable space and A ⊂ S. Then u ∈ cl(A) iff there is a sequence of points of A that converges to u (in the topology of S). Separation Axioms. It should be noted that a sequence can be divergent and still have accumulation points. For example {2, 0, 3/2, −1/2, 4/3, −2/3, . . . } does not converge but has both 1 and −1 as accumulation points. In arbitrary topological spaces, limit points of sequences are in general not unique. For example, in the trivial topology of S any sequence converges to all points of S. In order to avoid such situations several separation axioms have been introduced, of which the three most important ones will be mentioned. 1.1.13 Definition. A topological space S is called Hausdorff if each two distinct points have disjoint neighborhoods (i.e., with empty intersection). The space S is called regular if it is Hausdorff and if each closed set and point not in this set have disjoint neighborhoods. Similarly, S is called normal if it is Hausdorff and if each two disjoint closed sets have disjoint neighborhoods. Most standard spaces that we meet in geometry and analysis are normal. The discrete topology on any set is normal, but the trivial topology is not even Hausdorff. It turns out that “Hausdorff” is the necessary and sufficient condition for uniqueness of limit points of sequences in first countable spaces (see Exercise 1.1-5). Since in Hausdorff space single points are closed (Exercise 1.1-6), we have the implications: normal =⇒ regular =⇒ Hausdorff. Counterexamples for each of the converses of these implications are given in Exercises 1.1-9 and 1.1-10. 1.1.14 Proposition.

A regular second-countable space is normal.

Proof. Let A and B be two disjoint closed sets in S. By regularity, for every point p ∈ A there are disjoint open neighborhoods Up of p and UB of B. Hence cl(Up ) ∩ B = ∅. Since { Up | p ∈ A } is an open covering of A, by the Lindel¨ of Lemma 1.1.6, there is a countable collection { Uk | k = 0, 1, 2, . . . } covering A. Thus S k≥1 Uk ⊃ A and cl(Uk ) ∩ B = ∅. S Similarly, find a family {Vk } such that k≥0 Vk ⊃ B and cl(Vk ) ∩ A = ∅. Then the sets Gn defined inductively by G0 = U0 and [

Gn+1 = Un+1 \

cl(Vk ),

k=0,1,...,n

are open and G =

S

n≥0

Gn ⊃ A, H =

S

n≥0

H n = Vn \

[

cl(Uk )

k=0,1,...,n

Hn ⊃ B are also open and disjoint.



In the remainder of this book, Euclidean n-space Rn will be understood to have the standard topology unless explicitly stated to the contrary. Some Additional Set Theory. For technical completeness we shall present the axiom of choice and an equivalent result. These notions will be used occasionally in the text, but can be skipped on a first reading.

8

1. Topology

Axiom of choice.

If S is a collection of nonempty sets, then there is a function [ χ:S→ S S∈S

such that χ(S) ∈ S for every S ∈ S. The function χ, called a choice function, chooses one element from each S ∈ S. Even though this statement seems self-evident, it has been shown to be equivalent to a number of nontrivial statements, using other axioms of set theory. To discuss them, we need a few definitions. An order on a set A is a binary relation, usually denoted by “≤” satisfying the following conditions: a≤a a ≤ b and b ≤ a implies a = b a ≤ b and b ≤ c implies a ≤ c

(reflexivity ), (antisymmetry ), and (transitivity ).

An ordered set A is called a chain if for every a, b ∈ A, a 6= b we have a ≤ b or b ≤ a. The set A is said to be well ordered if it is a chain and every nonempty subset B has a first element; i.e., there exists an element b ∈ B such that b ≤ x for all x ∈ B. An upper bound u ∈ A of a chain C ⊂ A is an element for which c ≤ u for all c ∈ C. A maximal element m of an ordered set A is an element for which there is no other a ∈ A such that m ≤ a, a 6= m; in other words x ≤ m for all x ∈ A that are comparable to m. We state the following without proof. Theorem. Given other axioms of set theory, the following statements are equivalent: (i) The axiom of choice. (ii) Product Axiom. If {Ai }i∈I is a collection of nonempty sets then the product space Y Ai = { (xi ) | xi ∈ Ai } i∈I

is nonempty. (iii) Zermelo’s Theorem. Any set can be well ordered. (iv) Zorn’s Theorem. If A is an ordered set for which every chain has an upper bound (i.e., A is inductively ordered ), then A has at least one maximal element.

Exercises  1.1-1.

Let A = { (x, y, z) ∈ R3 | 0 ≤ x < 1 and y 2 + z 2 ≤ 1 }. Find int(A).

 1.1-2.

Show that any finite set in Rn is closed.

 1.1-3.

Find the closure of the set { 1/n | n = 1, 2, . . . } in R.

 1.1-4.

Let A ⊂ R. Show that sup(A) ∈ cl(A) where sup(A) is the supremum (least upper bound) of A.

 1.1-5.

Show that a first countable space is Hausdorff iff all sequences have at most one limit point.

 1.1-6.

(i) Prove that in a Hausdorff space, single points are closed.

(ii) Prove that a topological space is Hausdorff iff the intersection of all closed neighborhoods of a point equals the point itself.  1.1-7.

Show that in a Hausdorff space S the following are equivalent;

(i) S is regular;

1.1 Topological Spaces

9

(ii) for every point p ∈ S and any of its neighborhoods U , there exists a closed neighborhood V of p such that V ⊂ U ; (iii) for any closed set A, the intersection of all of the closed neighborhoods of A equals A.  1.1-8. (i) Show that if V(p) denotes the set of all neighborhoods of a point p ∈ S, a topological space, then the following are satisfied: V1. if A ⊃ U and U ∈ V(p), then A ∈ V(p); V2. every finite intersection of elements in V(p) is an element of V(p); V3. p belongs to all elements of V(p); V4. if V ∈ V(p) then there is a set U ∈ V(p), U ⊂ V such that for all q ∈ U , U ∈ V(q). (ii) If for each p ∈ S there is a family V(p) of subsets of S satisfying V1–V4, prove that there is a unique topology O on S such that for each p ∈ S, the family V(p) is the set of neighborhoods of p in the topology O. Hint: Prove uniqueness first and then define elements of O as being subsets A ⊂ S satisfying: for each p ∈ A, we have A ∈ V(p).  1.1-9. Let S = { p = (x, y) ∈ R2 | y ≥ 0 } and denote the usual ε-disk about p in the plane R2 by Dε (p) = { q | kq − pk < e }, where ε > 0. Define ( Dε (p) ∩ S, if p = (x, y) with y > 0; Bε (p) = { (x, y) ∈ Dε (p) | y > 0 } ∪ {p}, if p = (x, 0). Prove the following: (i) V(p) = { U ⊂ S | there exists Bε (p) ⊂ U } satisfies V1–V4 of Exercise 1.1-8. Thus S becomes a topological space. (ii) S is first countable. (iii) S is Hausdorff. (iv) S is separable. Hint: The set { (x, y) ∈ S | x, y ∈ Q, y > 0 } is dense in S. (v) S is not second countable. Hint: Assume the contrary and get a contradiction by looking at the points (x, 0) of S. (vi) S is not regular. Hint: Try to separate the point (x0 , 0) from the set { (x, 0) | x ∈ R }\{(x0 , 0)}.  1.1-10.

With the same notations as in the preceding exercise, except changing Bε (p) to ( Dε (p) ∩ S, if p = (x, y) with y > 0; Bε (p) = Dε (x, ε) ∪ {p}, if p = (x, 0),

show that (i)–(v) of Exercise 1.1-9 remain valid and that (vi) S is regular; Hint: Use Exercise 1.1-7.

10

1. Topology

(vii) S is not normal. Hint: Try to separate the set { (x, 0) | x ∈ Q } from the set { (x, 0) | x ∈ R\Q }.  1.1-11. Prove the following properties of the boundary operation and show by example that each inclusion cannot be replaced by equality. Bd1. bd(A) = bd(S\A); Bd2. bd(bd(A) ⊂ bd(A); Bd3. bd(A ∪ B) ⊂ bd(A) ∪ bd(B) ⊂ bd(A ∪ B) ∪ A ∪ B; Bd4. bd(bd(bd(A))) = bd(bd(A)). Properties Bd1–Bd4 may be used to characterize the topology.  1.1-12. Let p be a (nonzero) polynomial in n variables z1 , . . . , zn with complex coefficients. Show that p−1 (0) has open dense complement. Hint: If p vanishes on an open set of Cn , then all its derivatives also vanish and hence all its coefficients are zero.  1.1-13. Show that a collection of subsets B of a set S is a basis for a topology O of S (that is, O consists of unions of elements of B and is a topology on S) if and only if the following three conditions hold: B1. ∅ ∈ B; B2. ∪B∈B B = S; B3. if B1 , B2 ∈ B, then B1 ∩ B2 is a union of elements of B.  1.1-14.

1.2

Show that every second countable space is separable.

Metric Spaces

One of the common ways to form a topological space is through the use of a distance function, also called a (topological) metric. For example, on Rn the standard distance d(x, y) =

n X

!21 2

(xi − yi )

i=1

between x = (x1 , . . . , xn ) and y = (y1 , . . . , yn ) can be used to construct the open disks and from them the topology. The abstraction of this proceeds as follows. 1.2.1 Definition. Let M be a set. A metric (also called a topological metric) on M is a function d : M × M → R such that for all m1 , m2 , m3 ∈ M, M1. d(m1 , m2 ) = 0 iff m1 = m2 (definiteness); M2. d(m1 , m2 ) = d(m2 , m1 ) (symmetry); and M3. d(m1 , m3 ) ≤ d(m1 , m2 ) + d(m2 , m3 ) (triangle inequality). A metric space is the pair (M, d); if there is no danger of confusion, just write M for (M, d). Taking m1 = m3 in M3 shows that d(m1 , m2 ) ≥ 0. It is proved in advanced calculus courses (and is geometrically clear) that the standard distance on Rn satisfies M1–M3.

1.2 Metric Spaces

The Topology on a Metric Space. 1.2.2 Definition.

11

The topology determined by a metric is defined as follows.

For ε > 0 and m ∈ M , the open ε-ball (or disk ) about m is defined by Dε (m) = { m0 ∈ M | d(m0 , m) < ε } ,

and the closed ε−ball is defined by Bε (m) = { m0 ∈ M | d(m0 , m) ≤ ε } . The collection of subsets of M that are unions of open disks defines the metric topology of the metric space (M, d). Two metrics on a set are called equivalent if they induce the same metric topology. 1.2.3 Proposition. (i) The open sets defined in the preceding definition is a topology. (ii) A set U ⊂ M is open iff for each m ∈ U there is an ε > 0 such that Dε (m) ⊂ U. Proof. To prove (i), first note that T1 and T3 are clearly satisfied. To prove T2, it suffices to show that the intersection of two disks is a union of disks, which in turn is implied by the fact that any point in the intersection of two disks sits in a smaller disk included in this intersection. To verify this, suppose that p ∈ Dε (m) ∩ Dδ (n) and let 0 < r < min(ε − d(p, m), δ − d(p, n)). Hence Dr (p) ⊂ Dε (m) ∩ Dδ (n), since for any x ∈ Dr (p), d(x, m) ≤ d(x, p) + d(p, m) < r + d(p, m) < ε, and similarly d(x, n) < δ. Next we turn to (ii). By definition of the metric topology, a set V is a neighborhood of m ∈ M iff there exists a disk Dε (m) ⊂ V. Thus the statement in the theorem is equivalent to U = int (U ).  Notice that every set M can be made into a metric space by the discrete metric defined by setting d(m, n) = 1 for all m 6= n. The metric topology of M is the discrete topology. Pseudometric Spaces. M3, and

A pseudometric on a set M is a function d : M × M → R that satisfies M2,

PM1. d(m, m) = 0 for all m. Thus the distance between distinct points can be zero for a pseudometric. The pseudometric topology is defined exactly as the metric space topology. Any set M can be made into a pseudometric space by the trivial pseudometric: d(m, n) = 0 for all m, n ∈ M ; the pseudometric topology on M is the trivial topology. Note that a pseudometric space is Hausdorff iff it is a metric space. Metric Spaces are Normal. To show that metric spaces are normal, it will be useful to have the notion of the distance from a point to a set. If M is a metric space (or pseudometric space) and u ∈ M , A ⊂ M, we define d(u, A) = inf { d(u, v) | v ∈ A } if A 6= ∅, and we set d(u, ∅) = ∞, so that on sets, d takes values in the extended reals. The diameter of a set A ⊂ M is defined by diam(A) = sup { d(u, v) | u, v ∈ A } . A set is called bounded if its diameter is finite. Clearly metric spaces are first-countable and Hausdorff; in fact: 1.2.4 Proposition. Proof.

Every metric space is normal.

Let A and B be closed, disjoint subsets of M , and let U = { u ∈ M | d(u, A) < d(u, B) } and V = { v ∈ M | d(v, A) > d(v, B) } .

It is verified that U and V are open, disjoint and A ⊂ U , B ⊂ V.



12

1. Topology

Completeness. We learn in calculus the importance of the notion of completeness of the real line. The general notion of a complete metric space is as follows. 1.2.5 Definition. Let M be a metric space with metric d and {un } a sequence in M . Then {un } is a Cauchy sequence if for all real ε > 0, there is an integer N such that n, m ≥ N implies d(un , um ) < ε. The space M is called complete if every Cauchy sequence converges. We claim that a sequence {un } converges to u iff for every ε > 0 there is an integer N such that n ≥ N implies d(un , u) < ε. This follows readily from the Definitions 1.1.11 and 1.2.2. We also claim that a convergent sequence {un } is a Cauchy sequence. To see this, let ε > 0 be given. Choose N such that n ≥ N implies d(un , u) < ε/2. Thus, n, m ≥ N implies d(un , um ) ≤ d(un , u) + d(u, um ) <

ε ε + =ε 2 2

by the triangle inequality. Completeness requires that, conversely, every Cauchy sequence converges. A basic fact about Rn is that with the standard metric, it is complete. The proof is found in any textbook on advanced calculus. Contraction Maps.

A key to many existence theorems in analysis is the following.

1.2.6 Theorem (Contraction Mapping Theorem). Let M be a complete metric space and f : M → M a mapping. Assume there is a constant k, where 0 ≤ k < 1 such that d(f (x), f (x)) ≤ k d(x, y), for all x, y ∈ M ; such an f is called a contraction. Then f has a unique fixed point; that is, there exists a unique x∗ ∈ M such that f (x∗ ) = x∗ . Proof. Let x0 be an arbitrary point of M and define recursively xi+1 = f (xi ), i = 0, 1, 2, . . . . Induction shows that d(xi , xi+1 ) ≤ k i d(x0 , x1 ). Using this repeatedly together with the triangle inequality shows that for i < j, d(xi , xj ) ≤ (k i + · · · + k j−1 ) d(x0 , x1 ). For 0 ≤ k < 1, 1 + k + k 2 + k 3 + . . . is a convergent series, and so k i + k i+1 + · · · + k j−1 → 0 as i, j → ∞. This shows that the sequence {xi } is Cauchy and thus by completeness of M it converges to a point x∗ . Since d(x∗ , f (x∗ )) ≤ d(x∗ , xi ) + d(xi , f (xi )) + d(f (xi ), f (x∗ )) ≤ (1 + k) d(x∗ , xi ) + k i d(x0 , x1 ) is arbitrarily small, it follows that x∗ = f (x∗ ), thus proving the existence of a fixed point of f . If x0 is another fixed point of f , then d(x0 , x∗ ) = d(f (x0 ), f (x∗ )) ≤ k d(x0 , x∗ ), which, by virtue of 0 ≤ k < 1, implies d(x0 , x∗ ) = 0, so x0 = x∗ . Thus, we have uniqueness.



The condition k < 1 is necessary, for if M = R and f (x) = x + 1, then k = 1, but f has no fixed point (see also Exercise 1.5-5). At this point the true significance of the contraction mapping theorem cannot be demonstrated. When applied to the right spaces, however, it will yield the inverse function theorem (Chapter 2) and the basic existence theorem for differential equations (Chapter 4). A hint of this is given in Exercise 1.2-9.

1.2 Metric Spaces

13

Exercises  1.2-1. Let d((x1 , y1 ), (x2 , y2 )) = sup(|x1 − x2 |, |y1 − y2 |). Show that d is a metric on R2 and is equivalent to the standard metric.  1.2-2.

Let f (x) = sin(1/x), x > 0. Find the distance between the graph of f and (0, 0).

 1.2-3.

Show that every separable metric space is second countable.

 1.2-4. Show that every metric space (M, d) with infinite diameter has an equivalent metric d1 such that the diameter of (M, d1 ) is equal to 1. Hint: Consider the new metric d1 (x, y) = d(x, y)/[1 + d(x, y)].  1.2-5. In a metric space M , let V(m) = { U ⊂ M | there exists ε > 0 such that Dε (m) ⊂ U }. Show that V(m) satisfies V1–V4 of Exercise 1.1-8. This shows how the metric topology can be defined in an alternative way starting from neighborhoods.  1.2-6.

In a metric space show that cl(A) = { u ∈ M | d(u, A) = 0 }.

Exercises 1.2-7–1.2-9 use the notion of continuity as well as integration from elementary calculus (see §1.3).  1.2-7.

Let M denote the set of continuous functions f : [0, 1] → R on the interval [0, 1]. Show that Z 1 d(f, g) = |f (x) − g(x)| dx 0

is a metric.  1.2-8.

Let M denote the set of all continuous functions f : [0, 1] → R. Set d(f, g) = sup { |f (x) − g(x)| | 0 ≤ x ≤ 1 }

(i) Show that d is a metric on M . (ii) Show that fn → f in M iff fn converges uniformly to f . (iii) By consulting theorems on uniform convergence from your advanced calculus text, show that M is a complete metric space.  1.2-9.

Let M be as in the previous exercise and define T : M → M by Z x T (f )(x) = a + K(x, y) f (y) dy, 0

where a is a constant and K is a continuous function of two variables. Let Z x  k = sup |K(x, y)|dy 0 ≤ x ≤ 1 0

and suppose k < 1. Prove the following: (i) T is a contraction. (ii) Deduce the existence of a unique solution of the integral equation Z x f (x) = a + K(x, y) f (y) dy. 0

(iii) Taking a special case of (ii), prove the “existence of ex .”  1.2-10.

Prove the following error estimate in the contraction mapping principle: d(xn , x∗ ) ≤

kn d(x0 , x1 ). 1−k

14

1. Topology

1.3

Continuity

Definition of Continuity. is as follows.

We learn about continuity in calculus. Its general setting in topological spaces

1.3.1 Definition. Let S and T be topological spaces and ϕ : S → T be a mapping. We say that ϕ is continuous at u ∈ S if for every neighborhood V of ϕ(u) there is a neighborhood U of u such that ϕ(U ) ⊂ V . If, for every open set V of T , ϕ−1 (V ) = { u ∈ S | ϕ(u) ∈ V } is open in S, ϕ is continuous. (Thus, ϕ is continuous if ϕ is continuous at each u ∈ S.) If the map ϕ : S → T is a bijection (i.e., one-to-one and onto), and both ϕ and ϕ−1 are continuous, ϕ is called a homeomorphism and S and T are said to be homeomorphic. For example, notice that any map from a discrete topological space to any topological space is continuous. Similarly, any map from an arbitrary topological space to the trivial topological space is continuous. Hence the identity map from the set S topologized with the discrete topology to S with the trivial topology is bijective and continuous, but its inverse is not continuous, hence it is not a homeomorphism. Properties of Continuous Maps. It follows from Definition 1.3.1, by taking complements and using the set theoretic identity S\ϕ−1 (A) = ϕ−1 (T \A), that ϕ : S → T is continuous iff the inverse image of every closed set is closed. Here are additional properties of continuous maps. Let S, T be topological spaces and ϕ : S → T. The following are equivalent:

1.3.2 Proposition. (i) ϕ is continuous;

(ii) ϕ(cl(A)) ⊂ cl(ϕ(A)) for every A ⊂ S; (iii) ϕ−1 (int(B)) ⊂ int(ϕ−1 (B)) for every B ⊂ T. Proof.

If ϕ is continuous, then ϕ−1 (cl(ϕ(A))) is closed. But A ⊂ ϕ−1 (cl(ϕ(A))),

and hence cl(A) ⊂ ϕ−1 (cl(ϕ(A))), that is, ϕ(cl(A)) ⊂ cl(ϕ(A)). Conversely, let B ⊂ T be closed and A = ϕ−1 (B). Then ϕ(cl(A)) ⊂ cl(ϕ(A)) = cl(B) = B; that is, cl(A) ⊂ ϕ−1 (B) = A, so A is closed. A similar argument shows that (ii) and (iii) are equivalent.



This proposition combined with Proposition 1.1.12 (or a direct argument) gives the following. 1.3.3 Corollary. Let S and T be topological spaces with S first countable and ϕ : S → T. The map ϕ is continuous iff for every sequence {un } converging to u, {ϕ(un )} converges to ϕ(u), for all u ∈ S. 1.3.4 Proposition.

The composition of two continuous maps is a continuous map.

Proof. If ϕ1 : S1 → S2 and ϕ2 : S2 → S3 are continuous maps and if U is open in S3 , then (ϕ2 ◦ϕ1 )−1 (U ) = −1 −1 ϕ−1 1 (ϕ2 (U )) is open in S1 since ϕ2 (U ) is open in S2 by continuity of ϕ2 and hence its inverse image by ϕ1 is open in S1 , by continuity of ϕ1 .  1.3.5 Corollary. composition.

The set of all homeomorphisms of a topological space to itself forms a group under

1.3 Continuity

15

Proof. Composition of maps is associative and has for identity element the identity mapping. Since the inverse of a homeomorphism is a homeomorphism by definition, and since for any two homeomorphisms −1 ϕ1 , ϕ2 of S to itself, the maps ϕ1 ◦ ϕ2 and (ϕ1 ◦ ϕ2 )−1 = ϕ−1 2 ◦ ϕ1 are continuous by Proposition 1.3.4, the corollary follows.  1.3.6 Proposition. The space of continuous maps f : S → R forms an algebra under pointwise addition and multiplication. That is, if f and g are continuous, then so are f + g and f g. Proof. that

Let s0 ∈ S be fixed and ε > 0. By continuity of f and g at s0 , there exists an open set U in S such |f (s) − f (s0 )| <

ε , 2

and |g(s) − g(s0 )| <

ε 2

for all s ∈ U. Then |(f + g)(s) − (f + g)(s0 )| ≤ |f (s) − f (s0 )| + |g(s) − g(s0 )| < ε. Similarly, for ε > 0, choose a neighborhood V of s0 such that |f (s) − f (s0 )| < δ,

|g(s) − g(s0 )| < δ

for all s ∈ V , where δ is any positive number satisfying (δ + |f (s0 )|)δ + |g(s0 )|δ < ε. Then |(f g)(s) − (f g)(s0 )| ≤ |(f (s)| |g(s) − g(s0 )| + |f (s) − f (s0 )| |g(s0 )| < (δ + |f (s0 )|)δ + δ|g(s0 )| < ε. Therefore, f + g and f g are continuous at s0 .



Open and Closed Maps. Continuity is defined by requiring that inverse images of open (closed) sets are open (closed). In many situations it is important to ask whether the image of an open (closed) set is open (closed). 1.3.7 Definition. A map ϕ : S → T, where S and T are topological spaces, is called open (resp., closed ) if the image of every open (resp., closed ) set in S is open (resp., closed ) in T . Thus, a homeomorphism is a bijective continuous open (closed) map. An example of an open map that is not closed is ϕ : ]0, 1[ → R,

x 7→ x,

the inclusion map. An example of a closed map that is not open is ϕ : R → R,

defined by x 7→ x2

which maps ]−1, 1[ to [0, 1[. An example of a map that is neither open nor closed is the map ϕ : ]−1, 1[→ R,

defined by x 7→ x2 .

Finally, note that the identity map of a set S topologized with the trivial and discrete topologies on the domain and range, respectively, is not continuous but is both open and closed.

16

1. Topology

Continuous Maps between Metric Spaces. of ε’s and δ’s familiar from calculus.

For these spaces, continuity may be expressed in terms

1.3.8 Proposition. Let (M1 , d1 ) and (M2 , d2 ) be metric spaces, and ϕ : M1 → M2 a given mapping. Then ϕ is continuous at u1 ∈ M1 iff for every ε > 0 there is a δ > 0 such that d1 (u1 , u01 ) < δ implies d2 (ϕ(u1 ), ϕ(u01 )) < ε. Proof. Let ϕ be continuous at u1 and consider Dε2 (ϕ(u1 )), the ε-disk at ϕ(u1 ) in M2 . Then there is a δ-disk Dδ1 (u1 ) in M1 such that ϕ(Dδ1 (u1 )) ⊂ Dε2 (ϕ(u1 )) by Definition 1.3.1; that is, d1 (u1 , u01 ) < δ implies d2 (ϕ(u1 ), ϕ(u01 )) < ε. Conversely, assume this latter condition is satisfied and let V be a neighborhood of ϕ(u1 ) in M2 . Choosing an ε-disk Dε2 (ϕ(u1 )) ⊂ V there exists δ > 0 such that ϕ(Dδ1 (u1 )) ⊂ Dε2 (ϕ(u1 )) by the foregoing argument. Thus ϕ is continuous at u1 .  Uniform Continuity and Convergence. nuity and uniform convergence.

In a metric space we also have the notions of uniform conti-

1.3.9 Definition. (i) Let (M1 , d1 ) and (M2 , d2 ) be metric spaces and ϕ : M1 → M2 . We say ϕ is uniformly continuous if for every ε > 0 there is a δ > 0 such that d1 (u, v) < δ implies d2 (ϕ(u), ϕ(v)) < ε. (ii) Let S be a set, M a metric space, ϕn : S → M , n = 1, 2, . . . , and ϕ : S → M be given mappings. We say ϕn → ϕ uniformly if for every ε > 0 there is an N such that d(ϕn (u), ϕ(u)) < ε for all n ≥ N and all u ∈ S. For example, a map satisfying d(ϕ(u), ϕ(v)) ≤ Kd(u, v) for a constant K is uniformly continuous. Uniform continuity and uniform convergence ideas come up in the construction of a metric on the space of continuous maps. This is considered next. 1.3.10 Proposition. Let M be a topological space and (N, d) be a complete metric space. Then the collection C(M, N ) of all bounded continuous maps ϕ : M → N forms a complete metric space with the metric d0 (ϕ, ψ) = sup{ d(ϕ(u), ψ(u)) | u ∈ M }. Proof. It is readily verified that d0 is a metric. Convergence of a sequence fn ∈ C(M, N ) to f ∈ C(M, N ) in the metric d0 is the same as uniform convergence, as is readily checked. (See Exercise 1.2-8.) Now, if {fn } is a Cauchy sequence in C(M, N ), then {fn (x)} is Cauchy for each x ∈ M since d(fn (x), fm (x)) ≤ d0 (fn , fm ). Thus fn converges pointwise, defining a function f (x). We must show that fn → f uniformly and that f is continuous. First, given ε > 0, choose N such that d0 (fn , fm ) < ε/2 if n, m ≥ N. Second, for any x ∈ M , pick Nx ≥ N so that ε d(fm (x), f (x)) < 2 if m ≥ Nx . Thus with n ≥ N and m ≥ Nx , d(fn (x), f (x)) ≤ d(fn (x), fm (x)) + d(fm (x), f (x)) <

ε ε + = ε, 2 2

so fn → f uniformly. The reader can similarly verify that f is continuous (see Exercise 1.3-6; look in any advanced calculus text such as Marsden and Hoffman [1993] for the case of Rn if you get stuck). 

1.4 Subspaces, Products, and Quotients

17

Exercises  1.3-1. Show that a map ϕ : S → T between the topological spaces S and T is continuous iff for every set B ⊂ T , cl(ϕ−1 (B)) ⊂ ϕ−1 (cl(B)). Show that continuity of ϕ need not not imply any inclusion relations between ϕ(int(A)) and int(ϕ(A)).  1.3-2.

Show that a map ϕ : S → T is continuous and closed if for every subset U ⊂ S, ϕ(cl(U )) = cl(ϕ(U )).

 1.3-3.

Show that compositions of open (closed) mappings are also open (closed) mappings.

 1.3-4.

Show that ϕ : ]0, ∞[ → ]0, ∞[ defined by ϕ(x) = 1/x is continuous but not uniformly continuous.

 1.3-5. Show that if d is a pseudometric on M , then the map d(·, A) : M → R, for A ⊂ M a fixed subset, is continuous.  1.3-6. If S is a topological space, T a metric space, and ϕn : S → T a sequence of continuous functions uniformly convergent to a mapping ϕ : S → T, then ϕ is continuous.  1.3-7. Consider [0, 2π) and S1 = {x ∈ R2 | kxk = 1} as subsets of R and R2 , respectively (with the usual topology induced from R and R2 , respectively). Define the map f : [0, 2π) → S1 ;

x 7→ (cos x, sin x).

Is f a homeomorphism? Why or why not?  1.3-8. Let S be a set. Is the identity map from S with the discrete topology to S with the trivial topology continuous? A homeomorphism?

1.4

Subspaces, Products, and Quotients

This section concerns the construction of new topological spaces from old ones. Subset Topology.

The first basic operation of this type we consider is the formation of subset topologies.

1.4.1 Definition. If A is a subset of a topological space S with topology O, the relative topology on A is defined by OA = {U ∩ A | U ∈ O} . In other words, the open subsets in A are declared to be those subsets that are intersections of open sets in S with A. The following identities show that OA is indeed a topology: (i) ∅ ∩ A = ∅, S ∩ A = A; (ii) (U1 ∩ A) ∩ (U2 ∩ A) = (U1 ∩ U2 ) ∩ A; and S S (iii) α (Uα ∩ A) = ( α Uα ) ∩ A. Example. The topology on the n − 1-dimensional sphere S n−1 = { x ∈ Rn | d(x, 0) = 1 } is the relative topology induced from Rn ; that is, a neighborhood of a point x ∈ S n−1 is a subset of S n−1 containing the set Dε (x) ∩ S n−1 for some ε > 0. Note that an open (closed) set in the relative topology of A is in general not open (closed) in S. For example, Dε (x) ∩ S n−1 is open in S n−1 but it is neither open nor closed in Rn . However, if A is open (closed) in S, then any open (closed) set in the relative topology is also open (closed) in S. If ϕ : S → T is a continuous mapping, then the restriction ϕ|A : A → T is also continuous in the relative topology. The converse is false. For example, the mapping ϕ : R → R defined by ϕ(x) = 0 if x ∈ Q and ϕ(x) = 1 if x ∈ R\Q is discontinuous, but ϕ|Q : Q → R is a constant mapping and is thus continuous.

18

1. Topology

Products.

We can build up larger spaces by taking products of given ones.

1.4.2 Definition.

Let S and T be topological spaces and S × T = { (u, v) | u ∈ S and v ∈ T }.

The product topology on S × T consists of all subsets that are unions of sets which have the form U × V , where U is open in S and V is open in T . Thus, sets of the form U × V , which are called open rectangles, form a basis for the topology. Products of more than two factors can be considered in a similar way; it is straightforward to verify that the map ((u, v), w) 7→ (u, (v, w)) is a homeomorphism of (S × T ) × Z onto S × (T × Z). Similarly, one sees that S × T is homeomorphic to T × S. Thus one can take products of any finite number of topological spaces and the factors can be grouped in any order; we simply write S1 × · · · × Sn for such a finite product. For example, Rn has the product topology of R × · · · × R (n times). Indeed, using the maximum metric d(x, y) = max (|xi − y i |), 1≤i≤n

which is equivalent to the standard one, we see that the ε-disk at x coincides with the set ]x1 − ε, x1 + ε[ × · · · × ]xn − ε, xn + ε[. For generalizations to infinite products see Exercise 1.4-11, and to metric spaces see Exercise 1.4-14. 1.4.3 Proposition. Let S and T be topological spaces and denote by p1 : S × T → S and p2 : S × T → T the canonical projections: p1 (s, t) = s and p2 (s, t) = t. Then (i) p1 and p2 are open mappings; and (ii) a mapping ϕ : X → S ×T, where X is a topological space, is continuous iff both the maps p1 ◦ϕ : X → S and p2 ◦ ϕ : X → T are continuous. Proof. Part (i) follows directly from the definitions. To prove (ii), note that ϕ is continuous iff ϕ−1 (U × V ) is open in X, for U ⊂ S and V ⊂ T open sets. Since ϕ−1 (U × V ) = ϕ−1 (U × T ) ∩ ϕ−1 (S × V ) = (p1 ◦ ϕ)−1 (U ) ∩ (p2 ◦ ϕ)−1 (V ), the assertion follows.



In general, the maps pi , i = 1, 2, are not closed. For example, if S = T = R the set A = { (x, y) | xy = 1, x > 0 } is closed in S × T = R2 , but p1 (A) = ]0, ∞[ which is not closed in S. 1.4.4 Proposition. A topological space S is Hausdorff iff the diagonal which is defined by ∆S = { (s, s) | s ∈ S } ⊂ S × S is a closed subspace of S × S, with the product topology. Proof. It is enough to remark that S is Hausdorff iff for every two distinct points p, q ∈ S there exist neighborhoods Up , Uq of p, q, respectively, such that (Up × Uq ) ∩ ∆S = ∅.  Quotient Spaces. In a number of places later in the book we are going to form new topological spaces by collapsing old ones. We define this process now and give some examples. 1.4.5 Definition. u, v, w ∈ S, (i) u ∼ u

Let S be a set. An equivalence relation ∼ on S is a binary relation such that for all

(reflexivity );

1.4 Subspaces, Products, and Quotients

(ii) u ∼ v iff v ∼ u

19

(symmetry ); and

(iii) u ∼ v and v ∼ w implies u ∼ w

(transitivity ).

The equivalence class containing u, denoted [u] , is defined by [u] = { v ∈ S | u ∼ v }. The set of equivalence classes is denoted S/∼, and the mapping π : S → S/∼ defined by u 7→ [u] is called the canonical projection. Note that S is the disjoint union of its equivalence classes. The collection of subsets U of S/∼ such that π −1 (U ) is open in S is a topology because (i) π −1 (∅) = ∅, π −1 (S/ ∼) = S; (ii) π −1 (U1 ∩ U2 ) = π −1 (U1 ) ∩ π −1 (U2 ); and S S (iii) π −1 ( α Uα ) = α π −1 (Uα ). 1.4.6 Definition. Let S be a topological space and ∼ an equivalence relation on S. Then the collection of sets { U ⊂ S/∼ | π −1 (U ) is open in S } is called the quotient topology on S/∼. 1.4.7 Examples. A. The Torus.

Consider R2 and the relation ∼ defined by (a1 , a2 ) ∼ (b1 , b2 )

if a1 − b1 ∈ Z and a2 − b2 ∈ Z

(Z denotes the integers). Then T2 = R2 /∼ is called the 2-torus. In addition to the quotient topology, it inherits a group structure by setting [(a1 , a2 )] + [(b1 , b2 )] = [(a1 , a2 ) + (b1 , b2 )]. The n-dimensional torus Tn is defined in a similar manner. The torus T2 may be obtained in two other ways. First, let  be the unit square in R2 with the subspace topology. Define ∼ by x ∼ y iff any of the following hold: (i) x = y; (ii) x1 = y1 , x2 = 0, y2 = 1; (iii) x1 = y1 , x2 = 1, y2 = 0; (iv) x2 = y2 , x1 = 0, y1 = 1; or (v) x2 = y2 , x1 = 1, y1 = 0, as indicated in Figure 1.4.1. Then T2 = /∼ . Second, define T2 = S 1 × S 1 , also shown in Figure 1.4.1. B. The Klein bottle. The Klein bottle is obtained by reversing one of the orientations on , as indicated in Figure 1.4.2. Then K = /∼ (the equivalence relation indicated) is the Klein bottle. Although it is realizable as a subset of R4 , it is convenient to picture it in R3 as shown. In a sense we will make precise in Chapter 6, one can show that K is not “orientable.” Also note that K does not inherit a group structure from R2 , as did T2 . C. Projective Space. On Rn \{0} define x ∼ y if there is a nonzero real constant λ such that x = λy. Then (Rn \{0})/∼ is called real projective (n − 1)-space and is denoted by RPn−1 . Alternatively, RPn−1 can be defined as S n−1 (the unit sphere in Rn ) with antipodal points x and −x identified. (It is easy to see that this gives a homeomorphic space.) One defines complex projective space CPn−1 in an analogous way where now λ is complex. 

20

1. Topology

Figure 1.4.1. A torus

Figure 1.4.2. A Klein bottle

Continuity of Maps on Quotients. space is continuous.

The following is a convenient way to tell when a map on a quotient

1.4.8 Proposition. Let ∼ be an equivalence relation on the topological space S and π : S → S/ ∼ the canonical projection. A map ϕ : S/ ∼ → T , where T is another topological space, is continuous iff ϕ ◦ π : S → T is continuous. Proof. ϕ is continuous iff for every open set V ⊂ T , ϕ−1 (V ) is open in S/∼, that is, iff the set (ϕ◦π)−1 (V ) is open in S.  1.4.9 Definition. The set Γ = { (s, s0 ) | s ∼ s0 } ⊂ S × S is called the graph of the equivalence relation ∼. The equivalence relation is called open (closed ) if the canonical projection π : S → S/∼ is open (closed ). We note that ∼ is open (closed) iff for any open (closed) subset A of S the set π −1 (π(A)) is open (closed). As in Proposition 1.4.8, for an open (closed) equivalence relation ∼ on S, a map ϕ : S/∼ → T is open (closed) iff ϕ ◦ π : S → T is open (closed). In particular, if ∼ is an open (closed) equivalence relation on S and ϕ : S/∼ → T is a bijective continuous map, then ϕ is a homeomorphism iff ϕ ◦ π is open (closed). 1.4.10 Proposition. If S/∼ is Hausdorff, then the graph Γ of ∼ is closed in S × S. If the equivalence relation ∼ is open and Γ is closed (as a subset of S × S), then S/∼ is Hausdorff.

1.4 Subspaces, Products, and Quotients

21

Proof. If S/∼ is Hausdorff, then ∆S/∼ is closed by Proposition 1.4.4 and hence Γ = (π × π)−1 (∆S/∼ ) is closed in S × S, where π × π : S × S → (S/∼) × (S/∼) is given by (π × π)(x, y) = ([x], [y]). Assume that Γ is closed and ∼ is open. If S/∼ is not Hausdorff then there are distinct points [x], [y] ∈ S/∼ such that for any pair of neighborhoods Ux and Uy of [x] and [y], respectively, we have Ux ∩ Uy 6= ∅. Let Vx and Vy be any open neighborhoods of x and y, respectively. Since ∼ is an open equivalence relation, π(Vx ) = Ux

and π(Vy ) = Uy

are open neighborhoods of [x] and [y] in S/∼. Since Ux ∩ Uy 6= ∅, there exist x0 ∈ Vx and y 0 ∈ Vy such that [x0 ] = [y 0 ]; that is, (x0 , y 0 ) ∈ Γ. Thus, (Vx × Vy ) ∩ Γ 6= ∅ for any neighborhoods Vx and Vy of x and y respectively and thus, (x, y) ∈ cl(Γ) by Proposition 1.1.9(i). Because Γ is closed, we see that (x, y) ∈ Γ, that is, [x] = [y], a contradiction. 

Exercises  1.4-1. Show that the sequence xn = 1/n in the topological space ]0, 1] (with the relative topology from R) does not converge.  1.4-2. If f : S → T is continuous and T is Hausdorff, show that the graph of f, Γf = { (s, f (s)) | s ∈ S } is closed in S × T .  1.4-3. Let X and Y be topological spaces with Y Hausdorff. Show that for any continuous maps f, g : X → Y , the set { x ∈ X | f (x) = g(x) } is closed in X. Conclude that if f (x) = g(x) at all points of a dense subset of X, then f = g. Hint: Consider the mapping x 7→ (f (x), g(x)) and use Proposition 1.4.4.  1.4-4. Define a topological manifold to be a space locally homeomorphic to Rn . Find a topological manifold that is not Hausdorff. Hint: Consider R with “extra origins.”  1.4-5. Show that a mapping f : S → T is continuous iff the mapping s 7→ (s, f (s)) of S to the graph Γf = { (s, f (s)) | s ∈ S } ⊂ S × T is a homeomorphism of S with Γf (give Γf the subspace topology induced from the product topology of S × T ).  1.4-6. Show that every subspace of a Hausdorff (resp., regular) space is Hausdorff (resp., regular). Conversely, if each point of a topological space has a closed neighborhood that is Hausdorff (resp., regular) in the subspace topology, then the topological space is Hausdorff (resp., regular). Hint: use Exercises 1.1-6 and 1.1-7.  1.4-7.

Show that a product of topological spaces is Hausdorff iff each factor is Hausdorff.

 1.4-8. Let S, T be topological spaces and ∼, ≈ be equivalence relations on S and T , respectively. Let ϕ : S → T be continuous such that s1 ∼ s2 implies ϕ(s1 ) ≈ ϕ(s2 ). Show that the induced mapping ϕˆ : S/∼ → T /≈ is continuous.  1.4-9. Let S be a Hausdorff space and assume there is a continuous map σ : S/ ∼ → S such that π ◦ σ = iS/∼ , the identity. Show that S/∼ is Hausdorff and σ(S/∼) is closed in S.  1.4-10. Let M and N be metric spaces, N complete, and ϕ : A → N be uniformly continuous (A with the induced metric topology). Show that ϕ has a unique extension ϕ : cl(A) → N that is uniformly continuous.  1.4-11. Let S be a set, Tα a family of topological spaces, and ϕα : S → Tα a family of mappings. Let B be the collection of finite intersections of sets of the form ϕ−1 α (Uα ) for Uα open in Tα . The initial topology on S given by the family ϕα : S → Tα has as basis the collection B. Show that this topology is characterized

22

1. Topology

by the fact that any mapping ϕ : R → S from a topological space R is continuous iff all ϕα ◦ ϕ : R → Tα are continuous. Show that the subspace and product topologies are initial topologies. Define the product of an arbitrary infinite family of topological spaces and describe the topology.  1.4-12. Let T be a set and ϕα : Sα → T a family of mappings, Sα topological spaces with topologies Oα . Let O = { U ⊂ T | ϕ−1 α (U ) ∈ O for each α }. Show that O is a topology on T , called the final topology on T given by the family ϕα : Sα → T . Show that this topology is characterized by the fact that any mapping ϕ : T → R is continuous iff ϕ ◦ ϕα : Sα → R are all continuous. Show that the quotient topology is a final topology.  1.4-13.

Show that in a complete metric space a subspace is closed iff it is complete.

 1.4-14. Show that a product of two metric spaces is also a metric space by finding at least three equivalent metrics. Show that the product is complete if each factor is complete.  1.4-15. Let fi : Xi → Yi , i = 1, 2 be maps between topological spaces. Define f1 × f2 : X1 × X2 → Y1 × Y2 by (f1 × f2 )(x1 , x2 ) := (f1 (x1 ), f2 (x2 )), for x1 ∈ X1 and x2 ∈ X2 . (i) Show that f1 × f2 is continuous if and only if both f1 and f2 are continuous. (ii) Show that f1 × f2 is open if and only if both f1 and f2 are open. (iii) Let X1 = X2 = Y1 = Y2 = R, f1 = id, the identity map on R and f2 = 0, the constant map equal to zero. Show that both f1 and f2 are closed but that the product map f1 × f2 is not closed.

1.5

Compactness

Some basic theorems of calculus, such as “every real valued continuous function on [a, b] attains its maximum and minimum” implicitly use the fact that [a, b] is compact. Definition of Compactness. The general definition of compactness is rather unintuitive at the beginning. In fact, the general formulation of compactness and the realization of it as a useful tool is one of the excellent achievements of topology. But one has to be patient to see the rewards of formulating the definition the way it is done. 1.5.1 Definition. SLet S be a topological space. Then S is called compact if for every covering of S by open sets Uα (i.e., α Uα = S) there is a finite subcollection of the Uα also covering S. A subset A ⊂ S is called compact if A is compact in the relative topology. A subset A is called relatively compact if cl(A) is compact. A space is called locally compact if it is Hausdorff and each point has a relatively compact neighborhood. Properties of Compactness. with some simple observations.

We shall soon see the true power of this notion, but let’s work up to this

1.5.2 Proposition. (i) If S is compact and A ⊂ S is closed, then A is compact. (ii) If ϕ : S → T is continuous and S is compact, then ϕ(S) is compact. Proof. To prove (i), let {Uα } be an open covering of A. Then {Uα , S\A} is an open covering of S and hence contains a finite subcollection of this covering also covering S. The elements of this finite collection, except S\A, cover A. To prove (ii), let {Uα } be an open covering of ϕ(S). Then {ϕ−1 (Uα )} is an open covering of S and thus, by compactness of S, a finite subcollection { ϕ−1 (Uα(i) ) | i = 1, . . . , n }, covers S. But then {Uα(i) }, i = 1, . . . , n covers ϕ(S) and thus ϕ(S) is compact. 

1.5 Compactness

23

In a Hausdorff space, compact subsets are closed (exercise). Thus if S is compact, T is Hausdorff and ϕ is continuous, then ϕ is closed; if ϕ is also bijective, then it is a homeomorphism. Compactness of Products. 1.5.3 Proposition.

It is a basic fact that the product of compact spaces is compact.

A product space S × T is compact iff both S and T are compact.

Proof. In view of Proposition 1.5.2 all we have to show is that if S and T are compact, so is S × T . Let {Aα } be a covering of S × T by open sets. Each Aα is the union of sets of the form U × V with U and V open in S and T , respectively. Let {Uβ × Vβ } be a covering of S × T by open rectangles. If we show that there exists a finite subcollection of Uβ × Vβ covering S × T , then clearly also a finite subcollection of {Aα } will cover S × T . A finite subcollection of {Uβ × Vβ } is found in the following way. Fix s ∈ S. Since the set {s} × T is compact, there is a finite collection Uβ1 × Vβ1 , . . . , Uβi(s) × Vβi(s) covering it. If Us =

T

j=1,...,i(s)

Uβj , then Us is open, contains s, and Us × Vβ1 , . . . , Us × Vβi(s)

covers {s} × T . Let Ws = Us × T ; then the collection {Ws } is an open covering of S × T . If we show that only a finite number of these Ws cover S × T , then since [ Ws = (Us × Vβj ), j=1,...,i(s)

it follows that a finite number of Uβ × Vβ will cover S × T . Now look at S × {t}, for t ∈ T fixed. Since this is compact, a finite subcollection Ws1 , . . . , Wsk covers it. But then [ Wsj = S × T, (1.5.1) j=1,...,k

which proves the result.



As we shall see shortly in Theorem 1.5.9, [−1, 1] is compact. Thus T1 is compact. It follows from Proposition 1.5.3 that the torus T2 , and inductively Tn , are compact. Thus, if π : R2 → T2 is the canonical projection we see that T2 is compact without R2 being compact; that is, the converse of Proposition 1.5.2(ii) is false. Nevertheless it sometimes occurs that one does have a converse; this leads to the notion of a proper map discussed in Exercise 1.5-10. Bolzano–Weierstrass Theorem.

This theorem links compactness with convergence of sequences.

1.5.4 Theorem (Bolzano–Weierstrass Theorem). If S is a compact first countable Hausdorff space, then every sequence has a convergent subsequence.1 The converse is also true in metric and second-countable Hausdorff spaces. Proof. Suppose S is compact and {un } contains no convergent subsequences. We may assume that the points of the sequence are distinct. Then cl({un }) = {un } is compact and, since S is first countable, each un has a neighborhood Un that contains no other um , for otherwise un would be a limit of a subsequence. Thus {Un } is an open covering of the compact subset {un } which contains no finite subcovering, a contradiction. 1 There

are compact Hausdorff spaces in which there are sequences with no convergent subsequences. See page 69 of Sims [1976] for more information. Also, there are non-compact spaces in which every sequence has a convergent subsequence (see Exercise 5Ec in Kelley [1975]).

24

1. Topology

Let S be second countable, Hausdorff, and such that every sequence has a convergent subsequence. If {Uα } is an open covering of S, by the Lindel¨of Lemma2 there is a countable collection { Un | n = 1, 2 . . . } also covering S. Thus we have to show that { Un | n = 1, 2, . . . } contains a finite collection covering S. If this is not the case, the family consisting of sets of the form [ S\ Ui i=1,...,n

consists of closed nonempty sets and has the property that [ [ S\ Ui ⊃ S\ i=1,...,n

Ui

i=1,...,m

for m ≥ n. Choose pn ∈ S\

[

Ui .

i=1,...,n

If { pn | n = 1, 2, . . . } is infinite, by hypothesis it contains a convergent subsequence; let its limit point be denoted p. Then [ p ∈ S\ Ui i=1,...,n

for all n, contradicting the fact that { Un | n = 1, 2, . . . } covers S. Thus, { pn | n = 1, 2, . . . } must be a finite set; that is, for all n ≥ N , pn = pN . But then again [ pN ∈ S\ Ui i=1,...,n

for all n, contradicting the fact that { Un | n = 1, 2, . . . } covers S. Hence S is compact. Let S be a metric space such that every sequence has a convergent subsequence. If we show that S is separable, then since S is a metric space it is second countable (Exercise 1.2-3), and by the preceding paragraph, it will be compact. Separability of S is proved in two steps. S First we show that for any ε > 0 there is a finite set of points {p1 , . . . , pn } such that S = i=1,...,n Dε (pi ). If this were false, there would exist an ε > 0 such that no finite number of ε-disks cover S. Let p1 ∈ S be arbitrary. Since Dε (p1 ) 6= S, there is a point p2 ∈ S\Dε (p1 ). Since Dε (p1 ) ∪ Dε (p2 ) 6= S, there is also a point p3 ∈ S\(Dε (p1 ) ∪ Dε (p2 )), etc. The sequence { pn | n = 1, 2, . . . } is infinite and d(pi , pj ) ≥ ε. But this sequence has a convergent subsequence by hypothesis, so this subsequence must be Cauchy, contradicting d(pi , pj ) ≥ ε for all i, j. Second, we show that the existence for every ε > 0 of a finite set {p1 , . . . , pn(ε) } such that [ S= Dε (pi ) i=1,...,n(ε)

implies S is separable. Let An denote this finite set for ε = 1/n and let [ A= An . n≥0

Thus A is countable and it is easily verified that cl(A) = S. 2 Note



that in the proof of the converse, second countability was used only to enable us to quote the Lindel¨ of Lemma. Thus, the converse holds for topological spaces X having the property that every open cover contains a countable subcover. Such spaces are called Lindel¨ of spaces.

1.5 Compactness

Total Boundedness.

25

A property that came up in the preceding proof turns out to be important.

1.5.5 Definition. Let S be a metric space. A subset A ⊂ S is called totally bounded if for any ε > 0 there exists a finite set {p1 , . . . , pn } in S such that [

A⊂

Dε (pi ).

i=1,...,n

1.5.6 Corollary. A metric space is compact iff it is complete and totally bounded. A subset of a complete metric space is relatively compact iff it is totally bounded. Proof. The previous proof shows that compactness implies total boundedness. As for compactness implying completeness, it is enough to remark that in this context, a Cauchy sequence having a convergent subsequence is itself convergent. Conversely, if S is complete and totally bounded, let { pn | n = 1, 2, . . . } be a sequence in S. By total boundedness, this sequence contains a Cauchy subsequence, which by completeness, converges. Thus S is compact by the Bolzano–Weierstrass theorem. The second statement now readily follows.  1.5.7 Proposition.

In a metric space compact sets are closed and bounded.

Proof. This is a particular case of the previous corollary but can be easily proved directly. If A is compact, it can be finitely covered by ε-disks: [ A= Dε (pi ). i=1,...,n

Thus, diam(A) ≤

n X

diam(Dε (pi )) = 2nε.



i=1

From Proposition 1.5.2 and Proposition 1.5.7, we conclude that 1.5.8 Corollary. inf.

If S is compact and ϕ : S → R is continuous, then ϕ is bounded and attains its sup and

Indeed, since S is compact, so is ϕ(S) and so ϕ(S) is closed and bounded. Thus (see Exercise 1.1-4) the inf and sup of this set are finite and are members of this set. Heine–Borel Theorem.

This result makes it easy to spot compactness in Euclidean spaces.

1.5.9 Theorem (Heine–Borel Theorem).

In Rn a closed and bounded set is compact.

Proof. By Proposition 1.5.2(i) it is enough to show that closed bounded rectangles are compact in Rn , which in turn is implied via Proposition 1.5.3 by the fact that closed bounded intervals are compact in R. To show that [−a, a], a > 0 is compact, it suffices to prove (by Corollary 1.5.6) that for any given ε > 0, [−a, a] can be finitely covered by intervals of the form ]p − ε, p + ε[, since we are accepting completeness of R. Let n be a positive integer such that a < nε. Let t ∈ [−a, a] and k be the largest (positive or negative) integer satisfying kε ≤ t. Then −n ≤ k ≤ n and kε ≤ t < (k + 1)ε. Thus any point t ∈ [−a, a] belongs to an interval of the form ]kε − ε, kε + ε[, where k = −n, . . . , 0, . . . , n and hence { ]kε − ε, kε + ε[ | k = 0, ±1, . . . , ±n } is a finite covering of [−a, a].  This theorem is also proved in virtually every textbook on advanced calculus.

26

1. Topology

Uniform Continuity. As is known from calculus, continuity of a function on an interval [a, b] implies uniform continuity. The generalization to metric spaces is the following. 1.5.10 Proposition. A continuous mapping ϕ : M1 → M2 , where M1 and M2 are metric spaces and M1 is compact, is uniformly continuous. Proof. The metrics on M1 and M2 are denoted by d1 and d2 . Fix ε > 0. Then for each p ∈ M1 , by continuity of ϕ there exists δp > 0 such that if d1 (p, q) < δp , then d2 (ϕ(p), ϕ(q)) < ε/2. Let Dδ1 /2 (p1 ), . . . , Dδn /2 (pn ) cover the compact space M1 and let δ = min{δ1 /2, . . . , δn /2}. Then if p, q ∈ M1 are such that d1 (p, q) < δ, there exists an index i, 1 ≤ i ≤ n, such that d1 (p, pi ) < δi /2 and thus d1 (pi , q) ≤ d1 (pi , p) + d1 (p, q) <

δi + δ ≤ δi . 2

Thus, d2 (ϕ(p), ϕ(q)) ≤ d2 (ϕ(p), ϕ(pi )) + d2 (ϕ(pi )), ϕ(q)) < ε.



Equicontinuity. A useful application of Corollary 1.5.6 concerns relatively compact sets in C(M, N ), for metric spaces (M, dM ) and (N, dN ) with M compact and N complete. Recall from §1.3 that we put a metric on C(M, N ) and that in this metric, convergence is the same as uniform convergence. 1.5.11 Definition. A subset F ⊂ C(M, N ) is called equicontinuous at m0 ∈ M , if given ε > 0, there exists δ > 0 such that whenever dM (m, m0 ) < δ, we have dN (ϕ(m), ϕ(m0 )) < ε for every ϕ ∈ F (δ is independent of ϕ). F is called equicontinuous, if it is equicontinuous at every point in M . 1.5.12 Theorem (Arzela–Ascoli Theorem). Let (M, dM ) and (N, dN ) be metric spaces, and assume that M is compact and N is complete. A set F ⊂ C(M, N ) is relatively compact iff it is equicontinuous and all the sets F(m) = { ϕ(m) | ϕ ∈ F } are relatively compact in N . Proof. If F is relatively compact, it is totally bounded and hence so are all the sets F(m). Since N is complete, by Corollary 1.5.6 the sets F(m) are relatively compact. Let {ϕ1 , . . . , ϕn } be the centers of the ε-disks covering F. Then there exists δ > 0 such that if dM (m, m0 ) < δ, we have dN (ϕi (m), ϕi (m0 )) ≤ ε/3, for i = 1, . . . , n and hence if ϕ ∈ F is arbitrary, ϕ lies in one of the ε-disks whose center, say, is ϕi , so that dN (ϕ(m), ϕ(m0 )) ≤ dN (ϕ(m), ϕi (m)) + dN (ϕi (m), ϕi (m0 )) +dN (ϕi (m0 ), ϕ(m0 )) < ε. This shows that F is equicontinuous. Conversely, since C(M, N ) is complete, by Corollary 1.5.6 we need only show that F is totally bounded. For ε > 0, find a neighborhood Um of m ∈ M such that for all m0 ∈ Um , dN (ϕ(m), ϕ(m0 )) < ε/4 for all ϕ ∈ F (this is possible by equicontinuity). Let Um(1) , . . . , Um(n) be a finite collection of these neighborhoods covering the compact space M . By assumption each F(m) is relatively compact, hence F(m(1)) ∪ · · · ∪ F(m(n)) is also relatively compact, and thus totally bounded. Let Dε/4 (x1 ), . . . , De/4 (xk ) cover this union. If A denotes the set of all mappings α : {1, . . . , n} → {1, . . . , k}, then A is finite and [ F= Fα , a∈A

where Fα = { ϕ ∈ F | dN (ϕ(m(i)), xα(i) ) < ε/4 for all i = 1, . . . , n }. But if ϕ, ψ ∈ Fα and m ∈ M , then m ∈ Dε/4 (xi ) for some i, and thus dN (ϕ(m), ψ(m)) ≤ dN (ϕ(m), ϕ(m(i)) + dN (ϕ(m(i)), xα(i) ) +dN (xα(i) , ψ(m(i))) + dN (ψ(m(i)), ψ(m)) < ε; that is, the diameter of Fα is ≤ ε, so F is totally bounded.



1.5 Compactness

27

Combining this with the Heine–Borel theorem, we get the following. 1.5.13 Corollary. If M is a compact metric space, a set F ⊂ C(M, Rn ) is relatively compact iff it is equicontinuous and uniformly bounded (i.e., kϕ(m)k ≤ constant for all ϕ ∈ F and m ∈ M ). The following example illustrates one way to use the Arzela–Ascoli theorem. Example. Let fn : [0, 1] → R be continuous and be such that |fn (x)| ≤ 100 and the derivatives fn0 exist and are uniformly bounded on ]0, 1[. Prove fn has a uniformly convergent subsequence. We verify that the set {fn } is equicontinuous and bounded. The hypothesis is that |fn0 (x)| ≤ M for a constant M . Thus by the mean-value theorem, |fn (x) − fn (y)| ≤ M |x − y|, so given ε we can choose δ = ε/M , independent of x, y, and n. Thus {fn } is equicontinuous. It is bounded because kfn k = sup |fn (x)| ≤ 100.  0≤x≤1

Exercises  1.5-1. Show that a topological space S is compact iff every family of closed subsets of S whose intersection is empty contains a finite subfamily whose intersection is empty.  1.5-2. Show that every compact metric space is separable. Hint: Use total boundedness.  1.5-3. Show that the space of Exercise 1.1-9 is not locally compact. Hint: Look at the sequence (1/n, 0).  1.5-4.

(i) Show that every closed subset of a locally compact space is locally compact.

(ii) Show that S × T is locally compact if both S and T are locally compact.  1.5-5. Let M be a compact metric space and T : M → M a map satisfying d(T (m1 ), T (m2 )) < d(m1 , m2 ) for m1 6= m2 . Show that T has a unique fixed point.  1.5-6. Let S be a compact topological space and ∼ an equivalence relation on S, so that S/∼ is compact. Prove that the following conditions are equivalent (cf. Proposition 1.4.10): (i) The graph C of ∼ is closed in S × S; (ii) ∼ is a closed equivalence relation; (iii) S/∼ is Hausdorff.  1.5-7. Let S be a Hausdorff space that is locally homeomorphic to a locally compact Hausdorff space (i.e., for each u ∈ S, there is a neighborhood of u homeomorphic, in the subspace topology, to an open subset of a locally compact Hausdorff space). Show that S is locally compact. In particular, Hausdorff spaces locally homeomorphic to Rn are locally compact. Is the conclusion true without the Hausdorff assumption?  1.5-8. Let M3 be the set of all 3 × 3 matrices with the topology obtained by regarding M3 as R9 . Let SO(3) = { A ∈ M3 | A is orthogonal and det A = 1 } . (i) Show that SO(3) is compact. (ii) Let P = { Q ∈ SO(3) | Q is symmetric } and let ϕ : RP2 → SO(3) be given by ϕ(`) = the rotation by π about the line ` ⊂ R3 . Show that ϕ maps the space RP2 homeomorphically onto P \{Identity}.

28

1. Topology

 1.5-9.

Let fn : [a, b] → R be uniformly bounded continuous functions. Set Z x Fn (x) = fn (t) dt, a ≤ x ≤ b. a

Prove that Fn has a uniformly convergent subsequence.  1.5-10. Let X and Y be topological spaces. A map f : X → Y is called proper if the inverse image of any compact set in Y is compact in X. (i) Show that the composition of two proper maps is again proper. (ii) Let X be a compact space, Y be a Hausdorff space and suppose that f : X → Y is a continuous map. Show that f is proper. (iii) Let f : X → Y be continuous and proper. If X is Hausdorff and Y is a locally compact space, show that X is also locally compact. Hint: If U is an open neighborhood of x ∈ X and K is a compact neighborhood of f (x) ∈ Y , then cl(U ) ∩ f −1 (K) is a closed subset of f −1 (K) and so is a compact neighborhood of x. (iv) Assume that f : X → Y is continuous, closed and such that f −1 (y) is compact for all y ∈ Y . Show that f is proper. Hint: Let K ⊂ Y be compact and {Uα }α∈A be an open covering of f −1 (K). Let F be the family of all finite subsets of A and define for every F ∈ F the open set UF = ∪α∈F Uα . For each y ∈ K the set f −1 (y) is compact so there is some F ∈ F such that f −1 (y) ⊂ UF , that is, y ∈ Y \f (X\UF ), which says that K is covered by the family of open sets {Y \f (X\UF ) | F ∈ F}. Thus there are finitely many such sets Y \f (X\UFi ), i = 1, . . . , n, whose union contains K. Thus, [ [ [ Uα . f −1 (K) ⊂ f −1 (Y \f (X\UFi )) ⊂ UFi = i=1,...,n

i=1,...,n

α∈F1

S

···

S

Fn

(v) Assume that X and Y are Hausdorff, Y is first countable, and X is either second countable or metric. Show that if f : X → Y is continuous and proper then it is closed. Conclude that under these hypotheses on X and Y , a continuous map f is proper if and only if it is closed and the inverse image of every point is compact. Hint: If f (an ) → y with an ∈ A, the set B = {y, f (an ) | n ∈ N} is compact. Therefore, A ∩ f −1 (B) is compact and {an }n∈N ⊂ A ∩ f −1 (B). By the Bolzano-Weierstrass Theorem, there is a convergent subsequence ank → a ∈ cl(A) = A. (vi) Assume that X and Y are Hausdorff, Y is first countable, and X is either second countable or metric. Show that a continuous map f : X → Y is proper if and only if the following condition holds: if {f (xn )}n∈N is a convergent sequence in Y there is a convergent subsequence {xnk }n∈N in X. Hint: If f is proper the argument in (v) shows that the stated property holds. Conversely the stated property immediately implies that f is closed and that f −1 (y) is compact by the Bolzano-Weierstrass Theorem, for every y ∈ Y . Now apply (iv).

1.6

Connectedness

Three types of connectedness treated in this section are arcwise connectedness, connectedness, and simple connectedness.

1.6 Connectedness

Arcwise Connectedness.

29

We begin with the most intuitive notion of connectedness.

1.6.1 Definition. Let S be a topological space and I = [0, 1] ⊂ R. An arc ϕ in S is a continuous mapping ϕ : I → S. If ϕ(0) = u, ϕ(1) = v, we say ϕ joins u and v; S is called arcwise connected if every two points in S can be joined by an arc in S. A space S is called locally arcwise connected if for each point x ∈ S and each neighborhood U of x, there is a neighborhood V of x such that any pair of points in V can be joined by an arc in U . For example, Rn is arcwise and locally arcwise connected: any two points of Rn can be joined by the straight line segment connecting them. A set A ⊂ Rn is called convex if this property holds for any two of its points. Thus, convex sets in Rn are arcwise and locally arcwise connected. A set with the trivial topology is arcwise and locally arcwise connected, but in the discrete topology it is neither (unless it has only one point). Connected Spaces.

Less intuitive is the basic notion of connectedness.

1.6.2 Definition. A topological space S is connected if ∅ and S are the only subsets of S that are both open and closed. A subset of S is connected if it is connected in the relative topology. A component A of S is a nonempty connected subset of S such that the only connected subset of S containing A is A itself; S is called locally connected if for every point x ∈ S and for any neighborhood V of x, there is a connected neighborhood U of x such that U ⊂ V . The components of a subset T ⊂ S are the components of T in the relative topology of T in S. For example, Rn and any convex subset of Rn are connected and locally connected. The union of two disjoint open convex sets is disconnected but is locally connected; its components are the two convex sets. The trivial topology is connected and locally connected, whereas the discrete topology is neither: its components are all the one-point sets. Connected spaces are characterized by the following. 1.6.3 Proposition.

The following are equivalent:

(i) S is not connected ; (ii) there is a nonempty proper subset of S that is both open and closed ; (iii) S is the disjoint union of two nonempty open sets; and (iv) S is the disjoint union of two nonempty closed sets. The sets in (iii) or (iv) are said to disconnect S. Proof. To prove that (i) implies (ii), assume there is a nonempty proper subset A of S that is both open and closed. Then S = A ∪ (S\A) with A, S\A open and nonempty. Conversely, if S = A ∪ B with A, B open and nonempty, then A is also closed, and thus A is a proper nonempty set of S that is both open and closed. The equivalences of the remaining assertions are similarly checked.  We leave it to the reader to show that a topological space S is locally connected if and only if the connected component of every open set is open. Behavior under Mappings.

Connectedness is preserved by continuous maps, as is shown next.

1.6.4 Proposition. If f : S → T is a continuous map of topological spaces and S is connected (resp., arcwise connected ) then so is f (S). Proof. Let S be arcwise connected and consider f (s1 ), f (s2 ) ∈ f (S) ⊂ T . If c : I → S, c(0) = s1 , c(1) = s2 is an arc connecting s1 to s2 , then clearly f ◦ c : I → T is an arc connecting f (s1 ) to f (s2 ); that is, f (S) is

30

1. Topology

arcwise connected. Let S be connected and assume f (S) ⊂ U ∪ V , where U and V are open and U ∩ V = ∅. Then f −1 (U ) and f −1 (V ) are open by continuity of f , f −1 (U ) ∪ f −1 (V ) = f −1 (U ∪ V ) ⊃ f −1 (f (S)) = S, and f −1 (U ) ∩ f −1 (V ) = f −1 (∅) = ∅, thus contradicting connectedness of S by Proposition 1.6.3. Hence f (S) is connected.  Arcwise Connected Spaces are Connected. 1.6.5 Lemma.

We shall use the following.

The only connected sets of R are the intervals (finite, infinite, open, closed, or half-open).

Proof. Let us prove that [a, b[ is connected; all other possibilities have identical proofs. If not, [a, b[ = U ∪V with U, V nonempty disjoint closed sets in [a, b[. Assume that a ∈ U . If x = sup(U ), then x ∈ U since U is closed in [a, b[, and x < b since V 6= ∅. But then ]x, b[ ⊂ V and, since V is closed, x ∈ V . Hence x ∈ U ∩ V , a contradiction. Conversely, let A be a connected set of R. We claim that [x, y] ⊂ A whenever x, y ∈ A, which implies that A is an interval. If not, there exists z ∈ [x, y] with z 6∈ A . But in this case ]−∞, z[ ∩ A and ]z, ∞[ ∩ A are open nonempty sets disconnecting A.  1.6.6 Proposition.

If S is arcwise connected then it is connected.

Proof. If not, there are nonempty, disjoint open sets U0 and U1 whose union is S. Let x0 ∈ U0 and x1 ∈ U1 and let ϕ be an arc joining x0 to x1 . Then V0 = ϕ−1 (U0 ) and V1 = ϕ−1 (U1 ) disconnect [0, 1].  A standard example of a space that is connected but is not arcwise connected nor locally connected, is 

(x, y) ∈ R2 | x > 0 and y = sin(1/x) ∪ { (0, y) | −1 < y < 1 } .

1.6.7 Proposition. If a space is connected and locally arcwise connected, it is arcwise connected. In particular, a space locally homeomorphic to Rn is connected iff it is arcwise connected. Proof.

Fix x ∈ S. The set A = { y ∈ S | y can be connected to x by an arc }

is nonempty and open since S is locally arcwise connected. For the same reason, S\A is open. Since S is connected we must have S\A = ∅; thus, A = S, that is, S is arcwise connected.  Intermediate Value Theorem. calculus.

Connectedness provides a general context for this theorem learned in

1.6.8 Theorem (Intermediate Value Theorem). Let S be a connected space and f : S → R be continuous. Then f assumes every value between any two values f (u) and f (v). Proof. Suppose f (u) < a < f (v) and f does not assume the value a. Then the set U = { u0 | f (u0 ) < a } is both open and closed in S.  An alternative proof uses the fact that f (S) is connected in R and therefore is an interval.

1.6 Connectedness

31

Miscellaneous Properties of Connectedness. 1.6.9 Proposition.

Let S be a topological space and B ⊂ S be connected.

(i) If B ⊂ A ⊂ cl(B), then A is connected. (ii) If Bα is a family of connected subsets of S and Bα ∩ B = 6 ∅, then ! [ B∪ Bα α

is connected. Proof. If A is not connected, A is the disjoint union of U1 ∩ A and U2 ∩ A where U1 and U2 are open in S. Then from Proposition 1.1.9(i), U1 ∩ B 6= ∅ and U2 ∩ B 6= ∅, so B is not connected. We leave (ii) as an exercise.  1.6.10 Corollary. The components of a topological space are closed. Also, S is the disjoint union of its components. If S is locally connected, the components are open as well as closed. 1.6.11 Proposition. Let S be a first countable compact Hausdorff space and {An } a sequence of closed, T connected subsets of S with An ⊂ An−1 . Then A = n≥1 An is connected. Proof. As S is normal, if A is not connected, A lies in two disjoint open subsets U1 and U2 of S. If An ∩ (S\U1 ) ∩ (S\U2 ) 6= ∅ for all n, then there is a sequence un ∈ An ∩ (S\U1 ) ∩ (S\U2 ) with a subsequence converging to u. As An , S\U1 , and S\U2 are closed sets, u ∈ A ∩ (S\U1 ) ∩ (S\U2 ), a contradiction. Hence some An is not connected.  Simple Connectivity.

This notion means, intuitively, that loops can be continuously shrunk to points.

1.6.12 Definition. Let S be a topological space and c : [0, 1] → S a continuous map such that c(0) = c(1) = p ∈ S. We call c a loop in S based at p. The loop c is called contractible if there is a continuous map H : [0, 1] × [0, 1] → S such that H(t, 0) = c(t) and H(0, s) = H(1, s) = H(t, 1) = p for all t ∈ [0, 1]. (See Figure 1.6.1.)

Figure 1.6.1. The loop c is contractible

We think of cs (t) = H(t, s) as a family of arcs connecting c0 = c to c1 , a constant arc; see Figure 1.6.1. Roughly speaking, a loop is contractible when it can be shrunk continuously to p by loops beginning and ending at p. The study of loops leads naturally to homotopy theory. In fact, the loops at p can, by successively traversing them, be made into a group called the fundamental group; see Exercise 1.6-6.

32

1. Topology

1.6.13 Definition.

A space S is simply connected if S is connected and every loop in S is contractible.

2

In the plane R there is an alternative approach to simple connectedness, by way of the Jordan curve theorem; namely, that every simple (nonintersecting) loop in R2 divides R2 (divides means that its complement has two components). The bounded component of the complement is called the interior, and a subset A of R2 is simply connected iff the interior of every loop in A lies in A. Alexandroff ’s Theorem. We close this section with an optional theorem sometimes used in Riemannian geometry (to show that a Riemannian manifold is second countable) that illustrates the interplay between various notions introduced in this chapter. 1.6.14 Theorem (Alexandroff’s Theorem). An arcwise connected locally compact metric space is separable and hence is second countable. Proof (Pfluger [1957]). Since the metric space M is locally compact, each m ∈ M has compact neighborhoods that are disks. Let m ∈ M and denote by r(m) the least upper bound of the radii of such disks. If r(m) = ∞, since every metric space is first countable, M can be written as a countable union of compact disks. But since each compact metric space is separable (Exercise 1.5-2), these disks and also their union will be separable, and so the theorem is proved in this case. If there is some m0 ∈ M such that r(m0 ) < ∞, then since r(m) ≤ r(m0 ) + d(m, m0 ), we see that r(m) < ∞ for all m ∈ M . By the preceding argument, if we show that M is a countable union of compact sets, the theorem is proved. Under these circumstances, second countability follows from Exercise 1.2-3. To show that M is a countable union of compact sets, define the set Gm by Gm = { m0 ∈ M | d(m0 , m) ≤ r(m)/2 }. These Gm are compact neighborhoods of m. Fix m(0) ∈ M and put A0 = Gm(0) , and, inductively, define An+1 =

[

{ Gm | m ∈ An }.

Since M is arcwise connected, every point m ∈ M can be connected by an arc to m(0), which in turn is covered by finitely many Gm . This shows that [ M= An . n≥0

Since A0 is compact, all that remains to be shown is that the other An are compact. Assume inductively that An is compact and let {m(i)} be an infinite sequence of points in An+1 . There exists m(i)0 ∈ An such that m(i) ∈ Gm(i)0 . Since An is assumed to be compact there is a subsequence m(ik )0 that converges to a point m0 ∈ An . But d(m(i), m0 ) ≤ d(m(i), m(i)0 ) + d(m(i)0 , m0 ) r(m(i)0 ) ≤ + d(m(i)0 , m0 ) 2 r(m0 ) 3d(m(i)0 , m0 ) ≤ + . 2 2 Hence for ik big enough, all m(ik ) are in the compact set { n ∈ M | d(n, m0 ) ≤ 3r(m0 )/2 }, so m(ik ) has a subsequence converging to a point m. The preceding inequality shows that m ∈ An+1 . By the Bolzano–Weierstrass theorem, An+1 is compact. 

1.7 Baire Spaces

33

Exercises  1.6-1. Let M be a topological space and H : M → R continuous. Suppose e ∈ int H(M ). Then show H −1 (e) divides M ; that is, M \H −1 (e) has at least two components.  1.6-2. Let O(3) be the set of orthogonal 3 × 3 matrices. Show that O(3) is not connected and that it has two components.  1.6-3. Show that S × T is connected (locally connected, arcwise connected, locally arcwise connected) iff both S and T are. Hint: For connectedness write [ S×T = [(S × {t}) ∪ ({s0 } × T )] t∈T

for s0 ∈ S fixed and use Proposition 1.6.9(ii).  1.6-4.

Show that S is locally connected iff every component of an open set is open.

 1.6-5. Show that the quotient space of a connected (locally connected, arcwise connected) space is also connected (locally connected, arcwise connected). Hint: For local connectedness use Exercise 1.6-4 and show that the inverse image by π of a component of an open set is a union of components.  1.6-6. (i) Let S and T be topological spaces. Two continuous maps f, g : T → S are called homotopic if there exists a continuous map F : [0, 1] × T → S such that F (0, t) = f (t) and F (1, t) = g(t) for all t ∈ T . Show that homotopy is an equivalence relation. (ii) Show that S is simply connected if and only if any two continuous paths c1 , c2 : [0, 1] → S satisfying c1 (0) = c2 (0), c1 (1) = c2 (1) are homotopic, via a homotopy which preserves the end points, that is, F (s, 0) = c1 (0) = c2 (0) and F (s, 1) = c1 (1) = c2 (1). (iii) Define the composition c1 ∗ c2 of two paths c1 , c2 : [0, 1] → S satisfying c1 (1) = c2 (0) by

(c1 ∗ c2 )(t) =

 c1 (2t)

if t ∈ [1, 1/2];

 c2 (2t − 1)

if t ∈ [1/2, 1].

Show that this composition, when defined, induces an associative operation on endpoints preserving homotopy classes of paths. (iv) Fix s0 ∈ S and consider the set π1 (S, s0 ) of endpoint fixing homotopy classes of paths starting and ending at s0 . Show that π1 (S, s0 ) is a group: the identity element is given by the class of the constant path equal to s0 and the inverse of c is given by the class of c(1 − t) . (v) Show that if S is arcwise connected, then π1 (S, s0 ) is isomorphic to π1 (S, s) for any s ∈ S. π1 (S) will denote any of these isomorphic groups. (vi) Show that if S is arcwise connected, then S is simply connected iff π1 (S) = 0.

1.7

Baire Spaces

The Baire condition on a topological space is fundamental to the idea of “genericity” in differential topology and dynamical systems; see Kelley [1975] and Choquet [1969] for additional information.

34

1. Topology

1.7.1 Definition. Let X be a topological space and A ⊂ X a subset. Then A is called residual if A is the intersection of a countable family of open dense subsets of X. A space X is called a Baire space if every residual set is dense. A set B ⊂ X is called a first category set if [ Cn , B⊂ n≥1

where Cn is closed with int(Cn ) = ∅. A second category set is a set which is not of the first category. A set B ⊂ X is called nowhere dense if int(cl(B)) = ∅, so that X\A is residual iff A is the union of a countable collection of nowhere dense closed sets, that is, iff A is of first category. Clearly, a countable intersection of residual sets is residual. In a Baire space X, if [ Cn , X= n≥1

where Cn are closed sets, then int(Cn ) 6= ∅ for some n. For if all int(Cn ) = ∅, then On = X\Cn are open, dense, and we have \ [ On = X\ Cn = ∅ n≥1

n≥1

contradicting the definition of Baire space. In other words, Baire spaces are of second category. 1.7.2 Proposition. space.

Let X be a Baire space. Then the closure of any open subset of X is also a Baire

Proof. Let U be an open subset of X and A ⊂ cl(U ) be residual in the relative topology of cl(U ); that is, A = ∩n∈N On , where On are open and dense in cl(U ). By the definition of the relative topology of cl(U ), there are open sets Dn in X such that On = cl(U ) ∩ Dn . For every x ∈ cl(U ), any open neighborhood of x in X has therefore a non-empty intersection with each Dn for all n ∈ N. Let En := Dn ∪ (X \ cl(U )). The set En is clearly open being the union of two open sets. We shall prove that En is also dense in X. To see this, we will show that if W is any open subset of X, then W ∩ En 6= ∅. There are two possibilities: W ∩ cl(U ) = ∅ or W ∩ cl(U ) 6= ∅. If W ∩ cl(U ) = ∅, then W ⊂ X \ cl(U ) and so W ⊂ En . If W ∩ cl(U ) 6= ∅, then by the remark above we have W ∩ En ⊃ W ∩ Dn 6= ∅ for all n ∈ N. This shows that the arbitrary open set W intesects all En , that is, En is an open and dense subset of X for all n ∈ N. Therefore ∩n∈N En is residual in X and is hence dense in X, since X is a Baire space by hypothesis. However, En ∩ cl(U ) = Dn ∩ cl(U ) = On so that ∩n∈N En ∩ cl(U ) = ∩n∈N On = A. Now let V be an open subset of X such that V ∩ cl(U ) 6= ∅. We wil show that V ∩ A 6= ∅ which will conclude the proof of the proposition. By definition of the closure of a set, we have U ∩ V 6= ∅ and since both U and V are open in X, so is U ∩ V . Density of ∩n∈N En in X implies that ∅ 6= ∩n∈N En ∩ U ∩ V ⊂ ∩n∈N En ∩ cl(U ) ∩ V = A ∩ V .  1.7.3 Proposition. Let X be a locally Baire space; that is, each point x ∈ X has a neighborhood U such that cl(U ) is a Baire space. Then X is a Baire space. Proof. By Proposition 1.7.2, it can be assumed that the neighborhoods Tappearing in the definition of a locally Baire space are open subset in X. Let A ⊂ X be residual, A = n≥1 On , where T cl(On ) = X . Then if U is an open set for which cl(U ) is a Baire space, from the equality A ∩ cl(U ) = n≥1 (On ∩ cl(U )) and the density of On ∩ cl(U ) in cl(U ) (if u ∈ cl(U ) and u ∈ O, O open in X, then O ∩ U 6= ∅, and therefore O ∩ U ∩ On 6= ∅), it follows that A ∩ cl(U ) is residual in cl(U ) hence dense in cl(U ), that is, cl(A) ∩ cl(U ) = cl(U ) so that cl(U ) ⊂ cl(A). Therefore X = cl(A).  1.7.4 Theorem (Baire Category Theorem). spaces.

Complete pseudometric and locally compact spaces are Baire

1.7 Baire Spaces

Proof.

35

Let X be a complete pseudometric space. Let U ⊂ X be open and \ On A= n≥1

be residual. We must show U ∩ A 6= ∅. Since cl(On ) = X, U ∩ On 6= ∅ for all n ≥ 1, and so we can choose a disk of diameter less than one, say V1 , such that cl(V1 ) ⊂ U ∩ O1 . Proceed inductively to obtain cl(Vn ) ⊂ U ∩ On ∩ Vn−1 , where Vn has diameter < 1/n. Let xn ∈ cl(Vn ). Clearly {xn } is a Cauchy sequence, and by completeness has a convergent subsequence with limit point x. Then \ x∈ cl(Vn ) n≥1

and so

 U ∩

 \

On  6= ∅;

n≥1

that is, A is dense in X. If X is a locally compact space the same proof works with the following modifications: Vn are chosen to be relatively compact open sets, and {xn } has a convergent subsequence since it lies in the compact set cl(V1 ).  To get a feeling for this theorem, let us proveTthat the set of rationals Q cannot be written as a countable intersection of open sets. For suppose Q = n≥1 On . Then each On is dense in R, since Q is, and so Cn = R\On is closed and nowhere dense. Since   [ R∪ Cn  n≥1

is a complete metric space (as well as a locally compact space), it is of second category, so Q or some Cn should have nonempty interior. But this is impossible. The notion of category can lead to interesting restrictions on a set. For example in a nondiscrete Hausdorff space, any countable set is first category since the one-point set is closed and nowhere dense. Hence in such a space every second category set is uncountable. In particular, nonfinite complete pseudometric and locally compact spaces are uncountable.

Exercises  1.7-1.

Let X be a Baire space. Show that

(i) X is a second category set; (ii) if U ⊂ X is open, then U is Baire.  1.7-2. Let X be a topological space. A set is called an Fσ if it is a countable union of closed sets, and is called a Gδ if it is a countable intersection of open sets. Prove that the following are equivalent: (i) X is a Baire space; (ii) any first category set in X has a dense complement;

36

1. Topology

(iii) the complement of every first category Fσ -set is a dense Gδ -set; (iv) for any countable family of closed sets {Cn } satisfying [ Cn , X= n≥1

the open set [

int(Cn )

n≥1

is dense in X. Hint: First show that (ii) is equivalent to (iv). SFor (ii) implies (iv), let Un = CnS\ int(Cn ) so that S n≥1 Un is a first category set and therefore X\ T n≥1 Un is dense and included in n≥1 int(Cn ). For the converse, assume X is not Baire so that A = n≥1 Un is not dense, even though all Un are open and dense. Then X = cl(A) ∪ { X\Un | n = 1, 2, . . . }. Put F0 = cl(A),

Fn = X\Un ,

and show that int(Fn ) = int(cl(A)) which is not dense.  1.7-3. Show that there is a residual set E in the metric space C([0, 1], R) such that each f ∈ E is not differentiable at any point. Do this by following the steps below. (i) Let Eε denote the set of all f ∈ C([0, 1], R) such that for every x ∈ [0, 1],   f (x + h) − f (x) ε diam 2 < |h| < ε > 1 h for ε > 0. Show that Eε is open and dense in C([0, 1], R). Hint: For any polynomial p ∈ C([0, 1], R), show that p + δ cos(kx) ∈ Eε for δ small and δk large. T (ii) Show that E = n≥1 E1/n is dense in C([0, 1], R). Hint: Use the Baire category theorem. (iii) Show that if f ∈ E, then f has no derivative at any point.  1.7-4. Prove that in a complete metric space (M, d) with no isolated points, no countable dense set is a Gδ -set. T Hint: Suppose E = {x1 , x2 , . . . } is dense in M and is also a Gδ set, that is, E = n>0 Vn with Vn open, n = 1, T2, . . . . Conclude that Vn is dense in M . Let Wn = Vn \{x1 , . . . , xn }. Show that Wn is dense in M and that n>0 Wn = ∅. This contradicts the Baire property.

Page 37

2 Banach Spaces and Differential Calculus

This version: January 8, 2007

Manifolds have enough structure to allow differentiation of maps between them. To set the stage for these concepts requires a development of differential calculus in linear spaces from a geometric point of view. The goal of this chapter is to provide this perspective. Perhaps the most important technical theorem for later use is the Implicit Function Theorem. A fairly detailed exposition of this topic will be given with examples chosen that are motivate by later needs in manifold theory. The basic language of tangents, the derivative as a linear map, and the chain rule, while elementary, are important for developing geometric and analytic skills needed in manifold theory. The main goal is to develop the theory of finite-dimensional manifolds. However, it is instructive and efficient to do the infinite-dimensional theory simultaneously. To avoid being sidetracked by infinite-dimensional technicalities at this stage, some functional analysis background and other topics special to the infinitedimensional case are presented in supplements. With this arrangement, readers who wish to concentrate on the finite-dimensional theory can do so with a minimum of distraction.

2.1

Banach Spaces

It is assumed the reader is familiar with the concepts of real and complex vector spaces. Banach spaces are vector spaces with the additional structure of a norm that defines a complete metric space. While most of this book is concerned with finite-dimensional spaces, much of the theory is really no harder in the general case, and the infinite-dimensional case is needed for certain applications. Thus, it makes sense to work in the setting of Banach spaces. In addition, although the primary concern is with real Banach spaces, the basic concepts needed for complex Banach spaces are introduced with little extra effort. Normed Spaces. We begin with the notion of a normed space; that is, a space in which one has a length measure for vectors. 2.1.1 Definition. A norm on a real ( resp. complex ) vector space E is a mapping from E into the real numbers, k · k : E → R; e 7→ kek, such that N1. kek ≥ 0 for all e ∈ E and kek = 0 implies e = 0 (positive definiteness);

38

2. Banach Spaces and Differential Calculus

N2. kλek = |λ| kek for all e ∈ E and λ ∈ R (resp. C) (homogeneity); N3. ke1 + e2 k ≤ ke1 k + ke2 k for all e1 , e2 ∈ E (triangle inequality). The pair (E, k · k) is called a normed space. If there is no danger of confusion, we sometimes just say “E is a normed space.” To distinguish different norms, different notations are sometimes used, for example, k · kE , k · k1 , ||| · |||, etc., for the norm. Note that N2 implies that the length of the zero vector is zero: k0k = 0. Example.

Euclidean space Rn with the standard, or Euclidean, norm p kxk = ((x1 )2 + · · · + (xn )2 ),

where x = (x1 , . . . , xn ) ∈ Rn , is a normed space. Proving that this norm satisfies the triangle inequality is probably easiest to do using properties of the inner product, which are considered below. Another norm on the same space is given by n X |||x||| = |xi |, i=1

as may be verified directly.



The triangle inequality N3 has the following important consequence: | ke1 k − ke2 k | ≤ ke1 − e2 k for all e1 , e2 ∈ E, which is proved in the following way: ke2 k = ke1 + (e2 − e1 )k ≤ ke1 k + ke1 − e2 k, ke1 k = ke2 + (e1 − e2 )k ≤ ke2 k + ke1 − e2 k, so that both ke2 k − ke1 k and ke1 k − ke2 k are smaller than or equal to ke1 − e2 k. Seminormed Spaces.

If N1 in Definition 2.1.1 is replaced by

N10 . kek ≥ 0 for all e ∈ E, the mapping k · k : E → R is called a seminorm. For example, the function defined on R2 by k(x, y)k = |x| is a seminorm. Inner Product Spaces. These are spaces in which, roughly speaking, one can measure angles between vectors as well as their lengths. 2.1.2 Definition. An inner product on a real vector space E is a mapping h·, ·i : E × E → R, which we denote (e1 , e2 ) 7→ he1 , e2 i such that I1. he, e1 + e2 i = he, e1 i + he, e2 i; I2. he, αe1 i = α he, e1 i for all α ∈ R; I3. he1 , e2 i = he2 , e1 i; I4. he, ei ≥ 0 and he, ei = 0 iff e = 0.

2.1 Banach Spaces

39

The standard inner product on Rn is hx, yi =

n X

xi yi ,

i=1

and I1–I4 are readily checked. For vector spaces over the complex numbers, the definition is modified slightly as follows. 2.1.20 Definition. A complex inner product or a Hermitian inner product on a complex vector space E is a mapping h·, ·i : E × E → C such that the following conditions hold: CI1. he, e1 + e2 i = he, e1 i + he, e2 i; CI2. hαe, e1 i = α he, e1 i; CI3. he1 , e2 i = he2 , e1 i (so he, ei is real); CI4. he, ei ≥ 0 and he, ei = 0 iff e = 0. These properties are to hold for all e, e1 , e2 ∈ E and α ∈ C; z denotes the complex conjugate of the complex number z. Note that CI2 and CI3 imply that he1 , αe2 i = α he1 , e2 i . Properties CI1–CI3 are also known in the literature under the name sesquilinearity . As is customary, for a complex number z we shall denote by z+z z−z 1/2 Re z = , Im z = , |z| = (zz) 2 2i its real and imaginary parts and its absolute value. Let Cn = C × · · · × C be complex n-space whose points are denoted by z = (z 1 , . . . , z n ). The standard inner product on Cn is defined by n X hz, wi = z i wi , i=1

and CI1–CI4 are readily checked. Also Cn is a normed space with 2

kzk =

n X

2

|z i | .

i=1

In Rn or Cn , property N3 is a little harder to check directly. However, as we shall show in Proposition 2.1.4, N3 follows from I1–I4 or CI1–CI4. In a (real or complex) inner product space E, two vectors e1 , e2 ∈ E are called orthogonal and we write e1 ⊥ e2 provided he1 , e2 i = 0. For a subset A ⊂ E, the set A⊥ defined by A⊥ = { e ∈ E | he, xi = 0 for all x ∈ A } is called the orthogonal complement of A. Two sets A, B ⊂ E are called orthogonal and we write A ⊥ B if hA, Bi = 0; that is, e1 ⊥ e2 for all e1 ∈ A and e2 ∈ B. Cauchy–Schwartz Inequality. of lengths.

This inequality will be a critical way to estimate inner products in terms

2.1.3 Theorem (Cauchy–Schwartz Inequality).

In a (real or complex ) inner product space, 1/2

|he1 , e2 i| ≤ he1 , e1 i Equality holds iff e1 , e2 are linearly dependent.

1/2

he2 , e2 i

.

40

2. Banach Spaces and Differential Calculus

Proof.

It suffices to prove the complex case. If α, β ∈ C, then 2

2

0 ≤ hαe1 + βe2 , αe1 + βe2 i = |α| he1 , e1 i + 2 Re(αβ he1 , e2 i) + |β| he2 , e2 i . If we set α = he2 , e2 i, and β = − he1 , e2 i, then this becomes 2

2

2

0 ≤ he2 , e2 i he1 , e1 i − 2 he2 , e2 i |he1 , e2 i| + |he1 , e2 i| he2 , e2 i , and so 2

2

he2 , e2 i |he1 , e2 i| ≤ he2 , e2 i he1 , e1 i . If e2 = 0, equality results in the statement of the proposition and there is nothing to prove. If e2 6= 0, the term he2 , e2 i in the preceding inequality can be cancelled since he2 , e2 i > 0 by CI4. Taking square roots yields the statement of the proposition. Finally, equality results if and only if αe1 + βe2 = he2 , e2 i e1 − he1 , e2 i e2 = 0 by CI4.  1/2

2.1.4 Proposition. Let (E, h·, ·i) be a (real or complex ) inner product space and set kek = he, ei (E, k · k) is a normed space.

. Then

Proof. N1 and N2 are straightforward verifications. As for N3, the Cauchy–Schwartz inequality and the obvious inequality Re(he1 , e2 i) ≤ |he1 , e2 i| imply 2

2

2

2

ke1 + e2 k = ke1 k + 2 Re(he1 , e2 i) + ke2 k ≤ ke1 k + 2|he1 , e2 i| + ke2 k 2

2

2

2

≤ ke1 k + 2ke1 k ke2 k + ke2 k = (ke1 k + ke2 k)

Polarization and the Parallelogram Law. next. 2.1.5 Proposition.



Some other useful facts about inner products are given

Let (E, h·, ·i) be an inner product space and k · k the corresponding norm. Then

(i) (Polarization) 2

2

4 he1 , e2 i = ke1 + e2 k − ke1 − e2 k . for E real, while 2

2

2

2

4 he1 , e2 i = ke1 + e2 k − ke1 − e2 k + ike1 + ie2 k − ike1 − ie2 k , if E is complex. (ii) (Parallelogram law) 2

2

2

2

2ke1 k + 2ke2 k = ke1 + e2 k + ke1 − e2 k . Proof.

(i) In the complex case, we manipulate the right-hand side as follows 2

2

2

ke1 + e2 k − ke1 − e2 k + ike1 + ie2 k − ike1 − ie2 k 2

2

= ke1 k + 2 Re(he1 , e2 i) + ke2 k 2

2

− ke1 k + 2 Re(he1 , e2 i) − ke2 k 2

2

2

2

+ ike1 k + 2i Re(he1 , ie2 i) + ike2 k

− ike1 k + 2i Re(he1 , ie2 i) − ike2 k = 4 Re(he1 , e2 i) + 4i Re(−i he1 , e2 i) = 4 Re(he1 , e2 i) + 4i Im(he1 , e2 i) = 4 he1 , e2 i . The real case is proved in a similar way.

2

2.1 Banach Spaces

41

(ii) We manipulate the right hand side: 2

2

2

2

2

ke1 + e2 k + ke1 − e2 k = ke1 k + 2 Re(he1 , e2 i) + ke2 k + ke1 k 2

− 2 Re(he1 , e2 i) + ke2 k 2

2

= 2ke1 k + 2ke2 k .



Not all norms come from an inner product. For example, the norm |||x||| =

n X

|xi |

i=1

is not induced by any inner product since this norm fails to satisfy the parallelogram law (see Exercise 2.1-1 for a discussion). Normed Spaces are Metric Spaces. We have seen that inner product spaces are normed spaces. Now we show that normed spaces are metric spaces. 2.1.6 Proposition. Let (E, k · k) be a normed (resp. a seminormed ) space and define d(e1 , e2 ) = ke1 − e2 k. Then (E, d) is a metric (resp. pseudometric) space. Proof.

The only non-obvious verification is the triangle inequality for the metric. By N3, we have d(e1 , e3 ) = ke1 − e3 k = k(e1 − e2 ) + (e2 − e3 )k ≤ ke1 − e2 k + ke2 − e3 k = d(e1 , e2 ) + d(e2 , e3 ).



Thus we have the following hierarchy of generality: More General → inner product spaces

⊂

normed spaces

⊂

metric spaces

⊂

topological spaces

← More Special Since inner product and normed spaces are metric spaces, we can use the concepts from Chapter 1. In a (semi)normed space, N3 and N2 imply that the maps (e1 , e2 ) 7→ e1 + e2 , (α, e) 7→ αe of E × E → E and C × E → E, respectively, are continuous. Hence for e0 ∈ E, and α0 ∈ C (α0 6= 0) fixed, the mappings e 7→ e0 +e, e 7→ α0 e are homeomorphisms. Thus, U is a neighborhood of the origin iff e+U = { e+x | x ∈ U } is a neighborhood of e ∈ E. In other words, all the neighborhoods of e ∈ E are sets that contain translates of disks centered at the origin. This constitutes a complete description of the topology of a (semi)normed vector space (E, k · k) . Finally, note that the inequality |ke1 k − ke2 k| ≤ ke1 − e2 k implies that the (semi)norm is uniformly continuous on E. In inner product spaces, the Cauchy–Schwartz inequality implies the continuity of the inner product as a function of two variables. Banach and Hilbert Spaces.

Now we are ready to add the crucial assumption of completeness.

2.1.7 Definition. Let (E, k · k) be a normed space. If the corresponding metric d is complete, we say (E, k · k) is a Banach space. If (E, h·, ·i) is an inner product space whose corresponding metric is complete, we say (E, h·, ·i) is a Hilbert space.

42

2. Banach Spaces and Differential Calculus

For example, it is proven in books on advanced calculus that Rn is complete. Thus, Rn with the standard norm is a Banach space and with the standard inner product is a Hilbert space. Not only is the standard norm on Rn complete, but so is the nonstandard one |||x||| =

n X

|xi |.

i=1

To see this, it is enough to note that |||xn − x||| → 0 iff xin → xi in R. However, this nonstandard norm is equivalent to the standard one in the following sense. 2.1.8 Definition.

Two norms on a vector space E are equivalent if they induce the same topology on E.

2.1.9 Proposition. for all e ∈ E,

Two norms k · k and ||| · ||| on E are equivalent iff there is a constant M such that, 1 |||e||| ≤ kek ≤ M |||e|||. M

Proof.

Let Br1 (x) = { y ∈ E | ky − xk ≤ r } ,

Br2 (x) = { y ∈ E | |||y − x||| ≤ r }

denote the two closed disks of radius r centered at x ∈ E in the two metrics defined by the norms k · k and ||| · |||, respectively. Since neighborhoods of an arbitrary point are translates of neighborhoods of the origin, the two topologies are the same iff for every R > 0, there are constants M1 , M2 > 0 such that 2 1 2 BM (0) ⊂ BR (0) ⊂ BM (0). 1 2

The first inclusion says that if |||x||| ≤ M1 , then kxk ≤ R, that is, if |||x||| ≤ 1, then kxk ≤ R/M1 . Thus, if e 6= 0, then

e kek R

|||e||| = |||e||| ≤ M1 , that is, kek ≤ (R/M1 )|||e||| for all e ∈ E. Similarly, the second inclusion is equivalent to the assertion that (R/M2 )|||e||| ≤ kek for all e ∈ E. Thus the two topologies are the same iff there exist constants N1 > 0, N2 > 0 such that N1 |||e||| ≤ kek ≤ N2 |||e||| for all e ∈ E. Taking M = max(N2 , 1/N1 ) gives the statement of the proposition. Products of Normed Spaces.



If E and F are normed vector spaces, the map k·k:E×F→R

defined by k(e, e0 )k = kek + ke0 k is a norm on E × F inducing the product topology. Equivalent norms on E × F are (e, e0 ) 7→ max(kek, ke0 k)

and

2

2 1/2

(e, e0 ) 7→ (kek + ke0 k )

.

The normed vector space E × F is usually denoted by E ⊕ F and called the direct sum of E and F. Note that E ⊕ F is a Banach space iff both E and F are. These statements are readily checked.

2.1 Banach Spaces

43

Finite Dimensional Spaces. In the finite dimensional case equivalence and completeness are automatic, according to the following result. 2.1.10 Proposition.

Let E be a finite-dimensional real or complex vector space. Then

(i) there is a norm on E; (ii) all norms on E are equivalent; (iii) all norms on E are complete. Proof.

Let e1 , . . . , en denote a basis of E, where n is the dimension of E.

(i) A norm on E is given, for example, by |||e||| =

n X

|ai |,

where e =

i=1

n X

ai ei .

i=1

(ii) Let k · k be any other norm on E. If e=

n X

ai ei

and f =

i=1

n X

bi ei ,

i=1

the inequality | kek − kf k| ≤ ke − f k ≤

n X

|ai − bi | kei k

i=1

≤ max {kei k} |||(a1 , . . . , an ) − (b1 , . . . , bn )||| 1≤i≤n

shows that the map

n

X

i x ei ∈ [0, ∞[ (x , . . . , x ) ∈ R 7→

1

n

n

i=1

is continuous with respect to the ||| · |||-norm on Rn (use Cn in the complex case). Since the unit ball in the ||| · ||| norm, namely the set S = { x ∈ Rn | |||x||| = 1 }, is closed and bounded, it is compact. The restriction of this map to S is a continuous, strictly positive function, so it attains its minimum M1 and maximum M2 on S; that is,

n

X

i x ei ≤ M2 0 < M1 ≤

i=1

for all (x1 , . . . , xn ) ∈ Rn such that |||(x1 , . . . , xn )||| = 1. Thus, for all (x1 , . . . , xn ) ∈ Rn , we have

n

X

1 n i M1 |||(x , . . . , x )||| ≤ x ei ≤ M2 |||(x1 , . . . , xn )|||,

i=1 Pn that is, M1 |||e||| ≤ kek ≤ M2 |||e|||, where e = i=1 xi ei . Taking M = max(M2 , 1/M1 ), Proposition 2.1.9 shows that ||| · ||| and k · k are equivalent norms. (iii) It is enough to observe that (x1 , . . . , xn ) ∈ Rn 7→

n X

xi ei ∈ E

i=1

is a norm-preserving map (i.e., an isometry) between (Rn , ||| · |||) and (E, ||| · |||).



The unit spheres for the three common norms on R2 are shown in Figure 2.1 The foregoing proof shows that compactness of the unit sphere in a finite-dimensional space is crucial. This fact is exploited in the following supplement.

44

2. Banach Spaces and Differential Calculus

y

y

y

x

x

x

Figure 2.1.1. The unit spheres for various norms

Supplement 2.1A A Characterization of Finite-Dimensional Spaces 2.1.11 Proposition. unit disk is compact.

A normed vector space is finite dimensional iff it is locally compact iff the closed

Proof. If E is finite dimensional, every neighborhood of the origin contains a compact neighborhood, namely a closed disk (the closed disk is homeomorphic to a closed and bounded set in Rn , so is compact by the Heine-Borel theorem; see the proof of Proposition 2.1.10(ii) and (iii)). This shows that E is locally compact. Conversely, if the closed unit disk B1 (0) ⊂ E is compact, there is a finite covering of B1 (0) by open discs of radius 1/2, say { D1/2 (xi ) | i = 1, . . . , n }. Let F = span{x1 , . . . , xn }. Since F is finite dimensional, it is homeomorphic to Ck (or Rk ) for some k ≤ n, and thus complete. Being a complete subspace of the metric space (E, k · k), it is closed. We claim that F = E. If not, there would exist v ∈ E, v 6∈ F. Since F = cl(F), the number d = inf{ kv − ek | e ∈ F } is strictly positive. Let r > d > 0 be such that Br (v) ∩ F 6= ∅. The set Br (v) ∩ F is closed and bounded in the finite-dimensional space F, so is compact. Since inf{ kv − ek | e ∈ F } = inf{ kv − ek | e ∈ Br (v) ∩ F } and the continuous function defined by e ∈ Br (v) ∩ F 7→ kv − ek ∈ ]0, ∞[ attains its minimum, there is a point e0 ∈ Br (v) ∩ F such that d = kv − e0 k. But then there is a point xi such that

v − e0

1

− x i < ,

kv − e0 k 2 so that

d 1 kv − e0 k = 2 2 Since e0 + kv − e0 kxi ∈ F, we necessarily have that kv − e0 − kv − e0 kxi k ≥ d, which is a contradiction.  kv − e0 − kv − e0 kxi k <

2.1.12 Examples. A.

Let X be a set and F a normed vector space. Define the set   B(X, F) = f : X → F sup kf (x)k < ∞ . x∈X

Then B(X, F) is easily seen to be a normed vector space with respect to the sup-norm, kf k∞ = supx∈X kf (x)k.

2.1 Banach Spaces

45

This is the topology of uniform convergence. We prove that if F is complete, then B(X, F) is a Banach space. Let {fn } be a Cauchy sequence in B(X, F), that is, kfn − fm k∞ < ε for n, m ≥ N (ε). Since for each x ∈ X, kf (x)k ≤ kf k∞ , it follows that {fn (x)} is a Cauchy sequence in F, whose limit we denote by f (x). In the inequality kfn (x) − fm (x)k < ε for all n, m ≥ N (ε), let m → ∞ and get kfn (x) − f (x)k ≤ ε for all n ≥ N (ε) and all x ∈ X, that is, kfn − f k∞ ≤ ε for n ≥ N (ε). This shows that fn − f ∈ B(X, F), and hence f ∈ B(X, F), and also that kfn − f k∞ → 0 as n → ∞. As a particular case, one obtains the Banach space cb consisting of all bounded real (or complex) sequences {an } with the norm, also called the sup-norm, defined by k{an }k∞ = supn |an |. B.

If X is a topological space, the space CB(X, F) = { f : X → F | f is continuous, f ∈ B(X, F) }

is closed in B(X, F) since the uniform limit of continuous functions is continuous. Thus, if F is Banach, so is CB(X, F). In particular, if X is a compact topological space and F is a Banach space, then C(X, F) = { f : X → F | f continuous }, is a Banach space. For example, the vector space C([0, 1], R) = { f : [0, 1] → R | f is continuous } is a Banach space with the norm kf k∞ = sup{ |f (x)| | x ∈ [0, 1] }. C. (For readers with some knowledge of measure theory.) Consider the space of real valued square integrable functions defined on an interval [a, b] ⊂ R, that is, functions f that satisfy Z b 2 |f (x)| dx < ∞. a

The function Z k · k : f 7→

b

!1/2 2

|f (x)| dx a

is, strictly speaking, not a norm on this space; for example, if ( 0 for x 6= a, f (x) = 1 for x = a, then kf k = 0, but f 6= 0. However, k · k does become a norm if we identify functions which differ only on a set of measure zero in [a, b], that is, which are equal almost everywhere. The resulting vector space of equivalence classes [f ] will be denoted L2 ([a, b]) or simply as L2 [a, b]. With the norm of the equivalence class [f ] defined as !1/2 Z b 2 k[f ]k = |f (x)| dx , a

(where f is any chosen representative of the equivalence class), L2 ([a, b]) is an (infinite-dimensional) Banach space. The only nontrivial part of this assertion is the completeness; this is proved in books on measure

46

2. Banach Spaces and Differential Calculus

theory, such as Royden [1968]. As is customary, [f ] is denoted simply by f . In fact, L2 [a, b] is a Hilbert space with Z b f (x)g(x) dx. hf, gi = a

If we use square integrable complex-valued functions we get a complex Hilbert space L2 ([a, b], C) with Z hf, gi =

b

f (x)g(x) dx. a

D. The space Lp ([a, b]) may be defined for each real number p ≥ 1 in an analogous fashion to L2 [a, b]. Functions f : [a, b] → R satisfying Z b p |f (x)| dx < ∞ a

are considered equivalent if they agree almost everywhere. The space Lp ([a, b]) is then defined to be the vector space of equivalence classes of functions equal almost everywhere. The map

k · kp : L [a, b] → R

given by

!1/p

b

Z

p

[f ] →

p

|f (x)| dx a

defines a norm, called the Lp –norm, which makes Lp [a, b] into an (infinite-dimensional) Banach space. E. Denote by C([a, b], R) the set of continuous real valued functions on [a, b]. With the L1 -norm, C([a, b], R) is not a Banach space. For example, the sequence of continuous functions fn shown in Figure 2.1.2 is a Cauchy sequence in the L1 -norm on C([0, 1], R) but does not have a continuous limit function. On the other hand, with the sup norm kf k = sup |f (x)|, x∈[0,1]

C([0, 1], R) is complete, that is, it is a Banach space, as in Example B.

y 1

1 x

Figure 2.1.2. fn converges in L1 , but not in C.



2.1 Banach Spaces

47

Quotients. As in the case of both topological spaces and vector spaces, quotient spaces of normed vector spaces play a fundamental role. 2.1.13 Proposition. Let E be a normed vector space, F a closed subspace, E/F the quotient vector space,1 and π : E → E/F the canonical projection defined by π(e) = [e] = e + F ∈ E/F. (i) The mapping k · k : E/F → R defined by k[e]k = inf{ ke + vk | v ∈ F } is a norm on E/F. (ii) π is continuous and the topology on E/F defined by the norm coincides with the quotient topology. In particular, π is open. (iii) If E is a Banach space, so is E/F. Proof.

(i) Clearly k[e]k ≥ 0 for all [e] ∈ E/F and k[0]k = inf{ kvk | v ∈ F } = 0.

If k[e]k = 0, then there is a sequence {vn } ⊂ F such that lim ke + vn k = 0.

n→∞

Thus limn→∞ vn = −e and since F is closed, e ∈ F; that is, [e] = 0. Thus N1 is verified and the necessity of having F closed becomes apparent. N2 and N3 are straightforward verifications. (ii) Since k[e]k ≤ kek, it is obvious that limn→∞ en = e implies lim π(en ) = lim [en ] = [e]

n→∞

n→∞

and hence π is continuous. Translation by a fixed vector is a homeomorphism. Thus to show that the topology of E/F is the quotient topology, it suffices to show that if [0] ∈ U and π −1 (U ) is a neighborhood of zero in E, then U is a neighborhood of [0] in E/F. Since π −1 (U ) is a neighborhood of zero in E, there exists a disk Dr (0) ⊂ π −1 (U ). But then π(Dr (0)) ⊂ U and π(Dr (0)) = { [e] | e ∈ Dr (0) } = { [e] | k[e]k < r }, so that U is a neighborhood of [0] in E/F. (iii) Let {[en ]} be a Cauchy sequence in E/F. We may assume without loss of generality that k[en ] − [en+1 ]k ≤ n n 1/2 . Inductively, we find points e0n ∈ [en ] such that ke0n − e0n+1 k < 1/2 . Thus {e0n } is Cauchy in E so it converges to, say, e ∈ E. Continuity of π implies that limn→∞ [en ] = [e].  The codimension of F in E is defined to be the dimension of E/F. We say F is of finite codimension if E/F is finite dimensional. 2.1.14 Definition. The closed subspace F of the Banach space E is said to be split, or complemented , if there is a closed subspace G ⊂ E such that E = F ⊕ G. The relation between split subspaces and quotients is simple: the projection map of E to G induces, in a natural way, a Banach space isomorphism of E/F with G. We leave this as a verification for the reader. One should note, however, that the quotient E/F is defined independent of any choice of split subspace and that, accordingly, the choice of G is not unique. 1 This quotient is the same as the quotient in the sense discussed in Chapter 1, with the equivalence relation being u ∼ v iff u − v ∈ F, so that the equivalence class of u is the set u + F.

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2. Banach Spaces and Differential Calculus

Supplement 2.1B Split Subspaces Definition 2.1.14 implicitly asks that the topology of E coincide with the product topology of F ⊕ G. We shall show in Supplement 2.2C that this topological condition can be dropped; that is, the closed subspace F is split iff E is the algebraic direct sum of F and the closed subspace G. As we noted above, if E = F ⊕ G then G is isomorphic to E/F. However, F need not split for E/F to be a Banach space, as we proved in Proposition 2.1.13. In finite-dimensional spaces, any subspace is closed and splits; however, in infinite dimensions this is false. For example, let E = Lp (S 1 ) and let F = { f ∈ E | fˆ(n) = 0 for n < 0 }, where

Z π 1 fˆ(n) = f (θ)e−inθ dθ 2π −π is the nth Fourier coefficient of f . Then F is closed in E, splits in E for 1 < p < ∞ by a theorem of M. Riesz (Theorem 17.26 of Rudin [1966]) but does not split in E for p = 1 (Example 5.19 of Rudin [1973]). The same result holds if E = C 0 (S 1 , C) and F has the same definition. Another example worth mentioning is E = `∞ , the Banach space of all bounded sequences, and F = c0 , the subspace of `∞ consisting of all sequences convergent to zero. The subspace F = c0 is closed in E = `∞ , but does not split. However, c0 splits in any separable Banach space which contains it isomorphically as a closed subspace by a theorem of Sobczyk; see Veech [1971]. If every subspace of a Banach space is complemented, the space must be isomorphic to a Hilbert space by a result of Lindenstrauss and Tzafriri [1971]. Supplement 2.2B gives some general criteria useful in nonlinear analysis for a subspace to be split. But the simplest situation occurs in Hilbert spaces. 2.1.15 Proposition. If E is a Hilbert space and F a closed subspace, then E = F ⊕ F⊥ . Thus every closed subspace of a Hilbert space splits. The proof of this theorem is done in three steps, the first two being important results in their own rights. 2.1.16 Theorem (Minimal Norm Elements in Closed Convex Sets). is, x, y ∈ C and 0 ≤ t ≤ 1 implies tx + (1 − t)y ∈ C,

If C is a closed convex set in E, that

then there exists a unique e0 ∈ C such that √

ke0 k = inf{ kek | e ∈ C }.

Proof. Let d = inf{ kek | e ∈ C } . Then there exists a sequence {en } satisfying the inequality d ≤ ken k2 < d+1/n; hence ken k2 → d. Since (en +em )/2 ∈ C, C being convex, it follows that k(en +em )/2k2 ≥ d. By the parallelogram law,



e 2

e 2

en + em 2

en − em 2

m n

= 2

+ 2 −







2 2 2 2   d 1 d 1 1 1 1 < + + + −d= + ; 2 2n 2 2m 2 n m √ that is, {en } is a Cauchy sequence in E. Let limn→∞ en = e0 . Continuity of the norm implies that d = limn→∞ ken k = ke0 k, and so the existence of √ an element of minimum norm in C is proved. Finally, if f0 is such that ke0 k = kf0 k = d, the parallelogram law implies

2

2

e

e0 − f0 2

f0

0

− e0 + f0 ≤ d + d − d = 0;

= 2 2 +

2

2 2 2 2 2 that is, e0 = f0 .



2.1 Banach Spaces

49

2.1.17 Lemma. Let F ⊂ E, F 6= E be a closed subspace of E. Then there exists a nonzero element e0 ∈ E such that e0 ⊥ F. Proof. Let e ∈ E, e 6∈ F. The set e − F = { e − v | v ∈ F } is convex and closed, so by the preceding theorem, it contains a unique element e0 = e − v0 ∈ e − F (so that v0 ∈ F) of minimum norm. Since e 6∈ F, it follows that e0 6= 0. We shall prove that e0 ⊥ F. Since e0 is of minimal norm in e − F, for any v ∈ F and λ ∈ C (resp., R), we have ke0 k = ke − v0 k ≤ ke − v0 + λvk = ke0 + λvk, 2

that is, 2 Re(λ hv, e0 i) + |λ| kvk2 ≥ 0. If λ = a he0 , vi, a ∈ R, a 6= 0, this becomes a|hv, e0 i|2 (2 + akvk2 ) ≥ 0 for all v ∈ F and a ∈ R, a 6= 0. This forces hv, e0 i = 0 for all v ∈ F, for otherwise, if −2/kvk2 < a < 0, the preceding expression would be negative.  Proof of Proposition 2.1.15. F ⊕ F⊥ is a closed subspace. If

It is easy to see that F⊥ is closed (Exercise 2.1-4). We now show that

{en + e0n } ⊂ F ⊕ F⊥ ,

{en } ⊂ F,

{e0n } ⊂ F⊥ ,

the relation k(en + e0n ) − (em + e0m )k2 = ken − e0n k2 + kem − e0m k2 shows that {en + e0n } is Cauchy iff both {en } ⊂ F and {e0n } ⊂ F⊥ are Cauchy. Thus if {en + e0n } converges, then there exist e ∈ F, e0 ∈ F⊥ such that limn→∞ en = e, limn→∞ e0n = e0 . Thus lim (en + e0n ) = e + e0 ∈ F ⊕ F⊥ .

n→∞

If F ⊕ F⊥ 6= E, then by the previous lemma there exists e0 ∈ E, e0 ∈ 6 F ⊕ F⊥ , e0 6= 0, e0 ⊥ (F ⊕ F⊥ ). Hence e0 ∈ F⊥ and e0 ∈ F so that he0 , e0 i = ke0 k2 = 0; that is, e0 = 0, a contradiction. 

Exercises  2.1-1. Show that a normed space P is an inner product space iff the norm satisfies the parallelogram law. Conclude that if n ≥ 2, |||x||| = |xi | on Rn does not arise from an inner product. Hint: Use the polarization identities over R and C to guess the corresponding inner-products.  2.1-2. Let c0 be the space of real sequences {an } such that an → 0 as n → ∞. Show that c0 is a closed subspace of the space cb of bounded sequences (see Example 2.1.12A) and conclude that c0 is a Banach space.  2.1-3.

Let E1 be the set of all C 1 functions f : [0, 1] → R with the norm kf k = sup |f (x)| + sup |f 0 (x)|. x∈[0,1]

x∈[0,1]

(i) Prove that E1 is a Banach space. (ii) Let E0 be the space of C 0 maps f : [0, 1] → R, as in Example 2.1.12B. Show that the inclusion map E1 → E0 is compact; that is, the unit ball in E1 has compact closure in E0 . Hint: Use the Arzela–Ascoli theorem.

50

2. Banach Spaces and Differential Calculus

 2.1-4. Let (E, h·, ·i) be an inner product space and A, B subsets of E. Define the sum of A and B by A + B = { a + b | a ∈ A, b ∈ B }. Show that: (i) A ⊂ B implies B ⊥ ⊂ A⊥ ; (ii) A⊥ is a closed subspace of E; (iii) A⊥ = (cl(span(A)))⊥ , (A⊥ )⊥ = cl(span(A)); (iv) (A + B)⊥ = A⊥ ∩ B ⊥ ; and (v) (cl(span(A)) ∩ cl(span(B)))⊥ = A⊥ + B ⊥ (not necessarily a direct sum).  2.1-5. A sequence {en } ⊂ E, where E is an inner product space, is said to be weakly convergent to e ∈ E iff all the numerical sequences hv, en i converge to hv, ei for all v ∈ E. Let ( ) ∞ X 2 2 ` (C) = {an } an ∈ C and |an | < ∞ n=1

and put h{an } , {bn }i =

∞ X

an bn .

n=1

Show that: (i) in any inner product space, convergence implies weak convergence; (ii) `2 (C) is an inner product space; (iii) the sequence (1, 0, 0, . . . ), (0, 1, 0, . . . ), (0, 0, 1, . . . ), . . . is not convergent but is weakly convergent to 0 in `2 (C). Note: `2 (C) is in fact complete, so it is a Hilbert space. The ambitious reader can attempt a direct proof or consult a book on real analysis such as Royden [1968].  2.1-6. Show that absolutely convergent series is converP∞ a normed vector space is a Banach space iff Pevery ∞ gent. (A series n=1 xn is called absolutely convergent if n=1 kxn k converges.)  2.1-7. Let E be a Banach space and F1 ⊂ F2 ⊂ E be closed subspaces such that F2 splits in E. Show that F1 splits in E iff F1 splits in F2 .  2.1-8. Let F be closed in E of finite codimension. Show that if G is a subspace of E containing F, then G is closed.  2.1-9. Let E be a Hilbert space. A set {ei }i∈I is called orthonormal if hei , ej i = δij , the Kronecker delta. An orthonormal set {ei }i∈I is a Hilbert basis if cl(span{ei }i∈I ) = E. (i) Let {ei }i∈I be an orthonormal set and {ei(1) , . . . , ei(n) } be any finite subset. Show that n X

 e, ei(j) 2 ≤ kek2 j=1

for any e ∈ E. Hint:

e0 = e −

X

j=1,...,n

is orthogonal to all { ei(j) | j = 1, . . . , n }.

 e, ei(j) ei(j)

2.2 Linear and Multilinear Mappings

51

(ii) Deduce from (i) that for any positive integer n, the set { i ∈ I | |he, ei i| > 1/n } has at most nkek2 elements. Hence at most countably many i ∈ I satisfy he, ei i 6= 0, for any e ∈ E. (iii) Show that any Hilbert space has a Hilbert basis. Hint: Use Zorn’s lemma and Lemma 2.1.17.

 (iv) If {ei }i∈I is a Hilbert basis in E, e ∈ E, and {ei(j) } is the (at most countable) set such that e, ei(j) 6= 0, show that ∞ X

 e, ei(j) 2 = kek2 . j=1

Hint: If

e0 =

X



 e, ei(j) ei(j) ,

j=1,...,∞

show that hei , e − e0 i = 0

for all i ∈ I

and then use maximality of {ei }i∈I . (v) Show that E is separable iff any Hilbert basis is at most countable. Hint: For the “if” part, show that the set ( n ) X αn en αk = ak + ibk , where ak and bk are rational k=1

√ is dense in E. For the “only if” part, show that since kei − ej k2 = 2, the disks of radius 1/ 2 centered at ei are all disjoint. ) (vi) If E is a separable Hilbert space, it is algebraically isomorphic either with Cn or `2 (C) (Rn or `2 (R)), and the algebraic isomorphism can be chosen to be norm preserving.

2.2

Linear and Multilinear Mappings

This section deals with various aspects of linear and multilinear maps between Banach spaces. We begin with a study of continuity and go on to study spaces of continuous linear and multilinear maps and some related fundamental theorems of linear analysis. Continuity and Boundedness. and possessing a certain bound.

We begin by showing for a linear map, the equivalence of continuity

2.2.1 Proposition. Let A : E → F be a linear map of normed spaces. Then A is continuous if and only if there is a constant M > 0 such that kAekF ≤ M kekE Proof.

for all e ∈ E.

Continuity of A at e0 ∈ E means that for any r > 0, there exists ρ > 0 such that A(e0 + Bρ (0E )) ⊂ Ae0 + Br (0F )

(0E denotes the zero element in E and Bs (0E ) denotes the closed disk of radius s centered at the origin in E). Since A is linear, this is equivalent to: if kekE ≤ ρ, then kAekF ≤ r. If M = r/ρ, continuity of A is thus equivalent to the following: kekE ≤ 1 implies kAekF ≤ M , which in turn is the same as: there exists M > 0 such that kAekF ≤ M kekE , which is seen by choosing the vector e/kekE in the preceding implication. 

52

2. Banach Spaces and Differential Calculus

Because of this proposition one says that a continuous linear map is bounded . 2.2.2 Proposition. Proof.

If E is finite dimensional and A : E → F is linear, then A is continuous.

Let {e1 , . . . , en } be a basis for E. Letting M1 = max (kAe1 k, . . . , kAen k)

and setting e = a1 e1 + · · · + an en , we see that kAek = ka1 Ae1 + · · · + an Aen k ≤ |a1 | kAe1 k + · · · + |an | kAen k ≤ M1 (|a1 | + · · · + |an |). P i Since E is finite dimensional, all norms on it are equivalent. Since |||e||| = |a | is a norm, it follows that |||e||| ≤ Ckek for a constant C. Let M = M1 C and use Proposition 2.2.1 to conclude that A is continuous.  Operator Norm.

The bound on continuous linear maps suggests a norm for such maps.

2.2.3 Definition. If E and F are normed spaces and A : E → F is a continuous linear map, let the operator norm of A be defined by   kAek e ∈ E, e = 6 0 kAk = sup kek (which is finite by Proposition 2.2.1). Let L(E, F) denote the space of all continuous linear maps of E to F. If F = C (resp., R), then L(E, C) (resp., L(E, R)) is denoted by E∗ and is called the complex (resp., real ) dual space of E. (It will always be clear from the context whether L(E, F) or E∗ means the real or complex linear maps or dual space; in most of the work later in this book it will mean the real case.) A straightforward verification gives the following equivalent definitions of kAk: kAk = inf{ M > 0 | kAek ≤ M kek for all e ∈ E } = sup{ kAek | kek ≤ 1 } = sup{ kAek | kek = 1 }. In particular, kAek ≤ kAk kek . If A ∈ L(E, F) and B ∈ L(F, G), where E, F, and G are normed spaces, then k(B ◦ A)(e)k = kB(A(e))k ≤ kBk kAek ≤ kBk kAk kek, and so k(B ◦ A)k ≤ kBk kAk. Equality does not hold in general. A simple example is obtained by choosing E = F = G = R2 , A(x, y) = (x, 0), and B(x, y) = (0, y), so that B ◦ A = 0 and kAk = kBk = 1 . 2.2.4 Proposition.

L(E, F) with the norm just defined is a normed space. It is a Banach space if F is.

Proof. Clearly kAk ≥ 0 and k0k = 0. If kAk = 0, then for any e ∈ E, kAek ≤ kAk kek = 0, so that A = 0 and thus N1 (see Definition 2.1.1) is verified. N2 and N3 are also straightforward to check. Now let F be a Banach space and {An } ⊂ L(E, F) be a Cauchy sequence. Because of the inequality kAn e − Am ek ≤ kAn − Am k kek for each e ∈ E, the sequence {An e} is Cauchy in F and hence is convergent. Let Ae = limn→∞ An e. This defines a map A : E → F, which is evidently linear. It remains to be shown that A is continuous and kAn − Ak → 0. If ε > 0 is given, there exists a natural number N (ε)such that for all m, n ≥ N (ε) we have kAn −Am k < ε. If kek ≤ 1, this implies kAn e − Am ek < ε, and now letting m → ∞, it follows that kAn e − Aek ≤ ε for all e with kek ≤ 1. Thus An − A ∈ L(E, F), hence A ∈ L(E, F) and kAn − Ak ≤ ε for all n ≥ N (ε); that is, kAn − Ak → 0. 

2.2 Linear and Multilinear Mappings

53

If a sequence {An } converges to A in L(E, F) in the sense that kAn − Ak → 0,

if An → A

that is,

in the norm topology, we say An → A in norm. This phrase is necessary since other topologies on L(E, F) are possible. For example, we say that An → A strongly if An e → Ae for each e ∈ E. Since kAn e − Aek ≤ kAn − Ak kek, norm convergence implies strong convergence. The converse is false as the following example shows. Let ∞ ( ) X 2 2 E = ` (R) = {an } an < ∞ n=1

with inner product h{an } , {bn }i =

∞ X

an bn .

n=1

Let en = (0, . . . , 0, 1, 0, . . . ) ∈ E,

F = R,

and An = hen , ·i ∈ L(E, F),

th

where the √ 1 in en is in the n slot. The sequence {An } is not Cauchy in the operator norm since kAn − Am k = 2, but if e = {am }, An (e) = hen , ei = an → 0, that is, An → 0 strongly. If both E and F are finite dimensional, strong convergence implies norm convergence. (To see this, choose a basis e1 , . . . , en of E and note that strong convergence is equivalent to Ak ei → Aei as k → ∞ for i = 1, . . . , n. Hence maxi kAei k = |||A||| is a norm yielding strong convergence. But all norms are equivalent in finite dimensions.)

Supplement 2.2A Dual Spaces Riesz Representation Theorem. Recall from elementary linear algebra that the dual space of a finite dimensional vector space of dimension n also has dimension n and so the space and its dual are isomorphic. (Here, the dual space can be taken in either the algebraic sense or the sense in which we have defined it—they are equivalent in the finite dimensional case, which follows from what we did in §2.1.) For general Banach spaces this is no longer true. However, it is true for Hilbert space. 2.2.5 Theorem (Riesz Representation Theorem). Let E be a real (resp., complex ) Hilbert space. The map e 7→ h·, ei is a linear (resp., antilinear ) norm-preserving isomorphism of E with E∗ ; for short, E ∼ = E∗ . (A map A : E → F between complex vector spaces is called antilinear if we have the identities A(e + e0 ) = Ae + Ae0 , and A(αe) = αAe.) Proof. Let fe = h·, ei. Since the inner product is continuous and linear in each argument, fe ∈ E∗ . From the Cauchy-Schwarz inequality and the definition of the norm of a linear functional, we see that kfe k = kek. The map A : E → E∗ defined by Ae = fe is clearly linear (resp. antilinear), norm preserving, and thus injective. It remains to prove surjectivity. Let f ∈ E∗ and ker(f ) = { e ∈ E | f (e) = 0 }. ker(f ) is a closed subspace in E. If ker(f ) = E, then f = 0 and f = A(0) so there is nothing to prove. If ker(f ) 6= E, then by Lemma 2.1.17 there exists e 6= 0 such that e ⊥ ker(f ). Then we claim that f = A(f (e)e/kek2 ). Indeed, any v ∈ E can be written as v=v−

f (v) f (v) f (v) e+ e and v − e ∈ ker(f ). f (e) f (e) f (e)



54

2. Banach Spaces and Differential Calculus

Thus, in a real Hilbert space E every continuous linear function ` : E → R can be written `(e) = he0 , ei for some e0 ∈ E and k`k = ke0 k. In a general Banach space E we do not have such a concrete realization of E∗ . However, one should not always attempt to identify E and E∗ , even in finite dimensions. In fact, distinguishing these spaces is fundamental in tensor analysis. Reflexive Spaces.

We have a canonical map i : E → E∗∗ defined by i(e)(`) = `(e).

Pause and look again at this strange but natural formula: i(e) ∈ E∗∗ = (E∗ )∗ , so i(e) is applied to the element ` ∈ E∗ . If E is a Banach space, it is easy to check that i is norm preserving. Thus, i(E) is a Banach subspace of E∗∗ . One calls E reflexive if i is onto. Hilbert spaces are reflexive, by Theorem 2.2.5. For example, let V = L2 (Rn ) with inner product Z hf, gi = f (x)g(x) dx, Rn

and let α : L2 (Rn ) → R be a continuous linear functional. Then the Riesz representation theorem guarantees that there exists a unique g ∈ L2 (Rn ) such that Z α(f ) = g(x)f (x) dx = hg, f i Rn

for all f ∈ L2 (Rn ). In general, if E is not a Hilbert space and we wish to represent a linear functional α in the form of α(f ) = hg, f i , we must regard g as an element of the dual space E∗ . For example, let E = C0 (Ω, R), where Ω ⊂ Rn . Each x ∈ Ω defines a linear functional Ex : C0 (Ω, R) → R; f 7→ f (x). This linear functional cannot be represented in the form Ex (f ) = hg, f i and, indeed, is not continuous in the L2 norm. Nevertheless, it is customary and useful to write such linear maps as if h , i were the L2 inner product. Thus one writes, symbolically, Z δ(x − x0 )f (x) dx,

Ex0 (f ) = Ω

which defines the Dirac delta function at x0 ; that is, g(x) = δ(x − x0 ).

Linear Extension Theorem. require the following.

Next we shall discuss integration of vector valued functions. We shall

2.2.6 Theorem (Linear Extension Theorem).

Let E, F, and G be normed vector spaces where

(i) F ⊂ E; (ii) G is a Banach space; and (iii) T ∈ L(F, G). Then the closure cl(F) of F is a normed vector subspace of E and T can be uniquely extended to a map T ∈ L(cl(F), G). Moreover, we have the equality kT k = kT k.

2.2 Linear and Multilinear Mappings

55

Proof. The fact that cl(F) is a linear subspace of E is easily checked. Note that if T exists it is unique by continuity. Let us prove the existence of T . If e ∈ cl(F), we can write e = limn→∞ en , where en ∈ F, so that kT en − T em k ≤ kT k ken − em k, which shows that the sequence {T en } is Cauchy in the Banach space G. Let T e = limn→∞ T en . This limit is independent of the sequence {en }, for if e = lim e0n , then kT en − T e0n k ≤ kT k(ken − ek + ke − e0n k), which proves that limn→∞ (T en ) = limn→∞ (T e0n ). It is simple to check the linearity of T . Since T e = T e for e ∈ F (because e = limn→∞ e), T is an extension of T . Finally,



kT ek = lim (T en ) = lim kT en k ≤ kT k lim ken k = kT k kek n→∞

n→∞

n→∞

shows that T ∈ L(cl(F), G) and kT k ≤ kT k. The inequality kT k ≤ kT k is obvious since T extends T .



Integration of Banach Space Valued Functions. As an application of the preceding Theorem, we define a Banach space valued integral that will be of use later on. Fix the closed interval [a, b] ⊂ R and the Banach space E. A map f : [a, b] → E is called a step function if there exists a partition a = t0 < t1 < · · · < tn = b such that f is constant on each interval [ti , ti+1 [. Using the standard notion of a refinement of a partition, it is clear that the sum of two step functions and the scalar multiples of step functions are also step functions. Thus the set S([a, b], E) of step functions is a vector subspace of B([a, b], E), the Banach space of all bounded functions (see Example 2.1.12A). The integral of a step function f is defined by b

Z

f=

n−1 X

a

(ti+1 − ti )f (ti ).

i=0

It is easily verified that this definition is independent of the partition. Also note that

Z

b Z b

f ≤ kf k ≤ (b − a) kf k∞ ,

a a where kf k∞ = supa≤t≤b |f (t)|; that is, Z

b

: S([a, b], E) → E a

is continuous and linear. By the linear extension theorem, it extends to a continuous linear map Z b ∈ L(cl(S([a, b], E)), E). a

2.2.7 Definition.

The extended linear map

Note that

Rb a

is called the Cauchy–Bochner integral.

Z

b Z b

f ≤ kf k ≤ (b − a) kf k∞ .

a a

The usual properties of the integral such as Z b Z c Z f= f+ a

a

b

Z f

are easily verified since they clearly hold for step functions.

Z f =−

and

c

b

a

a

f b

56

2. Banach Spaces and Differential Calculus

The space cl(S([a, b], E) contains enough interesting functions for our purposes, namely C 0 ([a, b], E) ⊂ cl(S([a, b], E)) ⊂ B([a, b], E). The first inclusion is proved in the following way. Since [a, b] is compact, each f ∈ C 0 ([a, b], E) is uniformly continuous. For ε > 0, let δ > 0 be given by uniform continuity of f for ε/2. Then take a partition a = t0 < · · · < tn = b such that |ti+1 − t1 | < δ and define a step function g by g|[ti , ti+1 [ = f (ti ) . Then the ε-disk Dε (f ) in B([a, b], E) contains g. Finally, note that if E and F are Banach spaces, A ∈ L(E, F), and f ∈ cl(S([a, b], E)), we have A ◦ f ∈ cl(S([a, b], F)) since kA ◦ fn − A ◦ f k ≤ kAk kfn − f k∞ , where fn are step functions in E. Moreover, Z b

Z A◦f =A

a

!

b

f a

since this relation is obtained as the limit of the same (easily verified) relation for step functions. The reader versed in Riemann integration should notice that this integral for E = R is less general than the Riemann integral; that is, the Riemann integral exists also for functions outside of cl(S([a, b], R)). For purposes of this book, however, this integral will suffice. Multilinear Mappings.

If E1 , . . . , Ek and F are linear spaces, a map A : E1 × · · · × Ek → F

is called k-multilinear if A(e1 , . . . , ek ) is linear in each argument separately. Linearity in the first argument means that A(λe1 + µf1 , e2 , . . . , ek ) = λA(e1 , e2 , . . . , ek ) + µA(f1 , e2 , . . . , ek ). We shall study multilinear mappings in detail in our study of tensors. They also come up in the study of differentiation, and we shall require a few facts about them for that purpose. 2.2.8 Definition. The space of all continuous k-multilinear maps from E1 × · · · × Ek to F is denoted L(E1 , . . . , Ek ; F). If Ei = E, 1 ≤ i ≤ k, this space is denoted Lk (E, F). As in Definition 2.1.1, a k-multilinear map A is continuous if and only if there is an M > 0 such that kA(e1 , . . . , ek )k ≤ M ke1 k · · · kek k for all ei ∈ Ei , 1 ≤ i ≤ k. We set  kAk = sup

 kA(e1 , . . . , ek )k e1 , . . . , ek 6= 0 , ke1 k · · · kek k

which makes L(E1 , . . . , Ek ; F) into a normed space that is complete if F is. Again kAk can also be defined as kAk = inf{ M > 0 | kA(e1 , . . . , en )k ≤ M ke1 k · · · ken k } = sup{ kA(e1 , . . . , en )k | ke1 k ≤ 1, . . . , ken k ≤ 1 } = sup{ kA(e1 , . . . , en )k | ke1 k = · · · = ken k = 1 }. 2.2.9 Proposition.

There are (natural ) norm-preserving isomorphisms L(E1 , L(E2 , . . . , Ek ; F)) ∼ = L(E1 , . . . , Ek ; F) ∼ = L(E1 , . . . , Ek−1 ; L(Ek , F)) ∼ = L(Ei1 , . . . , Eik ; F)

where (i1 , . . . , ik ) is a permutation of (1, . . . , k).

2.2 Linear and Multilinear Mappings

Proof.

57

For A ∈ L(E1 , L(E2 , . . . , Ek ; F)), define A0 ∈ L(E1 , . . . , Ek ; F) by A0 (e1 , . . . , ek ) = A(e1 )(e2 , . . . , ek ).

The association A 7→ A0 is clearly linear and kA0 k = kAk. The other isomorphisms are proved similarly.  In a similar way, we can identify L(R, F) (or L(C, F) if F is complex) with F: to A ∈ L(R, F) we associate A(1) ∈ F; again kAk = kA(1)k. As a special case of Proposition 2.2.9 note that L(E, E∗ ) ∼ = L2 (E, R) (or 2 L (E; C), if E is complex). This isomorphism will be useful when we consider second derivatives. Permutations. We shall need a few facts about the permutation group on k elements. The information we cite is obtainable from virtually any elementary algebra book. The permutation group on k elements, denoted Sk , consists of all bijections σ : {1, . . . , k} → {1, . . . , k} together with the structure of a group under composition. Clearly, Sk has order k!, that is, Sk has k! elements. One of the more subtle but very useful properties of permutations is the notion of the sign of a permutation. The sign is a homomorphism sign : Sk → {−1, 1} , where {−1, 1} is the two element group under standard multiplication. We will define it shortly. Being a homomorphism means that for σ, τ ∈ Sk , sign(σ ◦ τ ) = (sign σ)(sign τ ). The kernel of “sign” consists of the subgroup of even permutations. Thus, a permutation σ is even when sign σ = +1 and is odd when sign σ = −1. The sign of a permutation is perhaps easiest to understand and define in terms of transpositions. A transposition is a permutation that swaps two elements of {1, . . . , k}, leaving the remainder fixed. It is a basic fact proved in algebra books (but it is intuitively obvious) that any permutation can be written as a product of transpositions. If it can be written as an even number of such transpositions, then its sign is defined to be +1, while its sign is −1 if it can be written as the product of an odd number of transpositions. It is not obvious that this gives a well defined definition independent of the way one writes the permutation as a product of transpositions, but this is proved in elementary books on group theory. The group Sk acts on the space Lk (E; F); that is, each σ ∈ Sk defines a map σ : Lk (E; F) → Lk (E; F) by (σA)(e1 , . . . , ek ) = A(eσ(1) , . . . , eσ(k) ). Note that (τ σ)A = τ (σA) for all τ, σ ∈ Sk . Accordingly, A ∈ Lk (E, F) is called symmetric (antisymmetric) if for any permutation σ ∈ Sk , σA = A (resp., σA = (sign σ)A.) 2.2.10 Definition. Let E and F be normed vector spaces. Let Lks (E; F) and Lka (E; F) denote the subspaces of symmetric and antisymmetric elements of Lk (E; F). Write S 0 (E, F) = F and S k (E, F) = { p : E → F | p(e) = A(e, . . . , e) for some A ∈ Lk (E; F) }. We call S k (E, F) the space of homogeneous polynomials of degree k from E to F. Note that Lks (E; F) and Lka (E; F) are closed in Lk (E; F); thus if F is a Banach space, so are Lks (E; F) and The antisymmetric maps Lka (E; F) will be studied in detail in Chapter 7. For technical purposes later in this chapter we will need a few facts about S k (E, F) which are given in the following supplement. Lka (E; F).

Supplement 2.2B Homogeneous Polynomials 2.2.11 Proposition.

58

2. Banach Spaces and Differential Calculus

(i) S k (E, F) is a normed vector space with respect to the following norm: kf k = inf{ M > 0 | kf (e)k ≤ M kekk } = sup{ kf (e)k | kek ≤ 1 } = sup{ kf (e)k | kek = 1 }. It is complete if F is. (ii) If f ∈ S k (E, F) and g ∈ S n (F, G), then g ◦ f ∈ S kn (E, G) and kg ◦ f k ≤ kgk kf k. (iii) (Polarization.) The mapping 0 : Lk (E, F) → S k (E, F) defined by A0 (e) = A(e, . . . , e) restricted to Lks (E; F) has an inverse `: S k (E, F) → Lks (E, F) given by 1 ∂k f (t1 e1 + · · · + tk ek ). `f (e1 , . . . , ek ) = k! ∂t1 · · · ∂tk t1 =···=tk =0 (Note that f (t1 e1 + · · · + tk ek ) is a polynomial in t1 , . . . , tk , so there is no problem in understanding what the derivatives on the right hand side mean.) (iv) For A ∈ Lk (E, F), kA0 k ≤ kAk ≤ (k k /k!)kA0 k, which implies the maps 0 and ` are continuous. Proof.

(i) and (ii) are proved exactly as for L(E, F) = S 1 (E, F).

(iii) For A ∈ Lks (E; F) one checks that A0 (t1 e1 + · · · + tk ek ) X k! a = ta1 · · · tj j A(e1 , . . . , e1 , . . . , ej , . . . , ej ), a1 ! · · · aj ! 1 a1 +···+aj =k

where each ei appears ai times, and ( 1, if k = j, 1 ∂k a ta1 · · · tj j = a1 ! · · · aj ! ∂t1 · · · ∂tk t1 =···=tk =0 1 0, if k 6= j. It follows that A(e1 , . . . , ek ) =

1 ∂k A0 (t1 e1 + · · · + tk ek ), k! ∂t1 · · · ∂tk

and for j 6= k, ∂j A0 (t1 e1 + · · · + tk ek ) = 0 ∂t1 · · · ∂tj t1 =···=tk =0 This means that`(A0 ) = A for any A ∈ Lks (E, F). Conversely, if f ∈ S k (E, F), then 1 ∂k (`f ) (e) =`f (e, . . . , e) = f (t1 e + · · · + tk e) k! ∂t1 · · · ∂tk t1 =···=tk =0 1 ∂k = (t1 + · · · + tk )k f (e) = f (e). k! ∂t1 · · · ∂tk t1 =···=tk =0 0

(iv) kA0 (e)k = kA(e, . . . , e)k ≤ kAk kekk , so kA0 k ≤ kAk. To prove the other inequality, note that if A ∈ Lks (E; F), then 1 X A(e1 , . . . , ek ) = ε1 · · · εk A0 (ε1 e1 + · · · + εk ek ), k!2k

2.2 Linear and Multilinear Mappings

59

where the sum is taken over all the 2k possibilities ε1 = ±1, . . . , εk = ±1. Put ke1 k = · · · = kek k = 1 and get kA0 (ε1 e1 + · · · + εk ek )k ≤ kA0 k kε1 e1 + · · · + εk ek kk ≤ kA0 k(|ε1 | ke1 k + · · · + |εk | kek k)k = kA0 kk k , whence kA(e1 , . . . , ek )k ≤

kk 0 kA k, k!

that is, kAk ≤

kk 0 kA k. k!



Let E = Rn , F = R, and e1 , . . . , en be the standard basis in Rn . For f ∈ S k (Rn , R), set ca1 ···an = f (e1 , . . . , e1 , . . . , en , . . . , en ), where each ei appears ai times, ai = 0, 1, . . . , k. If e = t1 e1 + · · · + tn en , the proof of (iii) shows that X

f (e) =`f (e, . . . , e) =

ca1 ...an ta1 1 · · · tann ,

a1 +···+an =k

that is, f is a homogeneous polynomial of degree k in t1 , . . . , tn in the usual algebraic sense. The constant k k /k! in (iv) is the best possible, as the following example shows. Write elements of Rk as x = (x1 , . . . , xk ) and introduce the norm |||(x1 , . . . , xk )||| = |x1 | + · · · + |xk |. Define A ∈ Lks (Rk , R) by A(x1 , . . . , xk ) =

1 X 1 xi1 . . . xkik , k!

where xi = (x1i , . . . , xki ) ∈ Rk and the sum is taken over all permutations of {1, . . . , k}. It is easily verified that kAk = 1/k! and kA0 k = 1/k k ; that is, kAk = (k k /k!)kA0 k. Thus, except for k = 1, the isomorphism 0 is not norm preserving. (This is a source of annoyance in the theory of formal power series and infinite-dimensional holomorphic mappings.)

Supplement 2.2C The Three Pillars of Linear Analysis The three fundamental theorems of linear analysis are the Hahn–Banach theorem, the open mapping theorem, and the uniform boundedness principle. See, for example, Banach [1932] and Riesz and Sz.-Nagy [1952] for further information. This supplement gives the classical proofs of these three fundamental theorems and derives some corollaries that will be used later. In finite dimensions these corollaries are all “obvious.”

60

2. Banach Spaces and Differential Calculus

Hahn–Banach Theorem.

This basic result guarantees a rich supply of continuous linear functionals.

2.2.12 Theorem (Hahn–Banach Theorem). Let E be a real or complex vector space, k · k : E → R a seminorm, and F ⊂ E a subspace. If f ∈ F∗ satisfies |f (e)| ≤ kek for all e ∈ F, then there exists a linear map f 0 : E → R (or C) such that f 0 |F = f and |f 0 (e)| ≤ kek for all e ∈ E. Proof. Real Case. First we show that f ∈ F∗ can be extended with the given property to F ⊕ span{e0 }, for a given e0 6∈ F. For e1 , e2 ∈ F we have f (e1 ) + f (e2 ) = f (e1 + e2 ) ≤ ke1 + e2 k ≤ ke1 + e0 k + ke2 − e0 k, so that f (e2 ) − ke2 − e0 k ≤ ke1 + e0 k − f (e1 ), and hence sup{ f (e2 ) − ke2 − e0 k | e2 ∈ F } ≤ inf{ ke1 + e0 k − f (e1 ) | e1 ∈ F }. Let a ∈ R be any number between the sup and inf in the preceding expression and define f 0 : F⊕ span {e0 } → R by f 0 (e+te0 ) = f (e)+ta. It is clear that f 0 is linear and that f 0 |F = f . To show that |f 0 (e+te0 )| ≤ ke+te0 k, note that by the definition of a, f (e2 ) − ke2 − e0 k ≤ a ≤ ke1 + e0 k − f (e1 ), so that by multiplying the second inequality by t ≥ 0 and the first by t < 0, we get the desired result. Second, one verifies that the set S = { (G, g) | F ⊂ G ⊂ E, G is a subspace of E, g ∈ G∗ , g|F = f, and |g(e)| ≤ kek for all e ∈ G } is inductively ordered with respect to the ordering (G1 , g1 ) ≤ (G2 , g2 ) iff

G1 ⊂ G2 and g2 |G1 = g1 .

Thus by Zorn’s lemma there exists a maximal element (F0 , f0 ) of S. Third, using the first step and the maximality of (F0 , f0 ), one concludes that F0 = E. Complex Case. Let f = Re f + i Im f and note that complex linearity implies that (Im f )(e) = −(Re f )(ie) for all e ∈ F. By the real case, Re f extends to a real linear continuous map (Re f )0 : E → R, such that |(Re f )0 (e)| ≤ kek for all e ∈ E. Define f 0 : E → C by f 0 (e) = (Re f )0 (e) − i(Re f )0 (ie) and note that f is complex linear and f 0 |F = f . To show that |f 0 (e)| ≤ kek for all e ∈ E, write f 0 (e) = |f 0 (e)| exp(iθ), so complex linearity of f 0 implies 0 f (e · exp(−iθ)) = |f 0 (e)| ∈ R, and hence |f 0 (e)| = f 0 (e · exp(−iθ)) = (Re f )0 (e · exp(−iθ)) ≤ ke · exp(−iθ)k = kek.



2.2.13 Corollary. Let (E, k · k) be a normed space, F ⊂ E a subspace, and f ∈ F∗ (the topological dual ). Then there exists f 0 ∈ E∗ such that f 0 |F = f and kf 0 k = kf k. Proof. We can assume f 6= 0. Then |||e||| = kf k kek is a norm on E and |f (e)| ≤ kf k kek = |||e||| for all e ∈ F . Applying the preceding theorem we get a linear map f 0 : E → R (or C) with the properties f 0 |F = f and |f 0 (e)| ≤ |||e||| for all e ∈ E. This says that kf 0 k ≤ kf k, and since f 0 extends f , it follows that kf k ≤ kf 0 k; that is, kf 0 k = kf k and f 0 ∈ E∗ .  Applying the corollary to the linear function ae 7→ a, for e ∈ E a fixed element, we get the following. 2.2.14 Corollary. Let E be a normed vector space and e 6= 0. Then there exists f ∈ E∗ such that f (e) 6= 0. In other words if f (e) = 0 for all f ∈ E∗ , then e = 0; that is, E∗ separates points of E.

2.2 Linear and Multilinear Mappings

Open Mapping Theorem.

61

This result states that surjective linear maps are open.

2.2.15 Theorem (Open Mapping Theorem of Banach–Schauder). suppose A ∈ L(E, F) is onto. Then A is an open mapping.

Let E and F be Banach spaces and

Proof. To show A is an open mapping, it suffices to prove that the set A(cl(D1 (0))) contains a disk centered at zero in F. Let r > 0. Since [ E= Dnr (0), n≥1

it follows that [

F=

(A(Dnr (0)))

n≥1

and hence [

cl(A(Dnr (0))) = F.

n≥1

Completeness of F implies that at least one of the sets cl(A(Dnr (0))) has a nonempty interior by the Baire category theorem 1.7.4. Because the mapping e ∈ E 7→ ne ∈ E is a homeomorphism, we conclude that cl(A(Dr (0))) contains some open set V ⊂ F. We shall prove that the origin of F is in int{cl[A(Dr (0))]} for some r > 0. Continuity of (e1 , e2 ) ∈ E × E 7→ e1 − e2 ∈ E assures the existence of an open set U ⊂ E such that U − U = { e1 − e2 | e1 , e2 ∈ U } ⊂ Dr (0). Choose s > 0 such that Ds (0) ⊂ U . Then cl(A(Dr (0))) ⊃ cl(A(U ) − A(U )) ⊃ cl(A(U )) − cl(A(U )) ⊃ cl (A (Ds (0))) − cl (A (Ds (0))) s s = (cl (A (Dr (0))) − cl (A (Dr (0)))) ⊃ (V − V ) . r r But V −V =

[

(V − e)

e∈V

is open and clearly contains 0 ∈ F. It follows that there exists a disk Dt (0) ⊂ F such that Dt (0) ⊂ cl(A(Dr (0))). P Now let ε(n) = 1/2n+1 , n = 0, 1, 2, . . . , so that 1 = n≥0 ε(n). By the foregoing result for each n there exists an η(n) > 0 such that Dη(n) (0) ⊂ cl(A(Dε(n) (0))). Clearly η(n) → 0. We shall prove that Dη(0) ⊂ A(cl(D1 (0))). For v ∈ Dη(0) (0) ⊂ cl(A(Dε(0) (0))) there exists e0 ∈ Dε(0) (0) such that kv − Ae0 k < η(1) and thus v − Ae0 ∈ cl(A(Dε(1) (0))), so there exists e1 ∈ Dε(1) (0) such that kv − Ae0 − Ae1 k < η(2), etc. Inductively one constructs a sequence en ∈ Dη(n) such that kv − Ae0 − · · · − Aen k < η(n + 1). The series P e is convergent because n n≥0

m

∞ m

X X X 1 1

, = 1, ei ≤

i+1 n+1

2 2 n=0 i=n+1 i=n+1 and E is complete. Let e =

P

n≥0 en

∈ E. Thus, Ae =

∞ X

Aen = v,

n=0

and kek ≤

∞ X n=0

ken k ≤

∞ X

1 = 1; n+1 2 n=0

62

2. Banach Spaces and Differential Calculus

that is, v ∈ Dη(0) (0) implies v = Ae, kek ≤ 1. Therefore, Dη(0) (0) ⊂ A(cl(D1 (0))).



An important consequence is the following. 2.2.16 Theorem (Banach’s Isomorphism Theorem). is a homeomorphism.

A continuous linear isomorphism of Banach spaces

Thus, if F and G are closed subspaces of the Banach space E and E is the algebraic direct sum of F and G, then the mapping (e, e0 ) ∈ F × G 7→ e + e0 ∈ E is a continuous isomorphism, and hence a homeomorphism; that is, E = F ⊕ G; this proves the comment at the beginning of Supplement 2.1B. Closed Graph Theorem.

This result characterizes continuity by closedness of the graph of a linear map.

2.2.17 Theorem (Closed Graph Theorem). A : E → F is continuous iff its graph

Suppose that E and F are Banach spaces. A linear map

ΓA = { (e, Ae) ∈ E × F | e ∈ E } is a closed subspace of E ⊕ F. Proof. It is readily verified that ΓA is a linear subspace of E ⊕ F. If A ∈ L(E, F), then ΓA is closed (see Exercise 1.4-2). Conversely, if ΓA is closed, then it is a Banach subspace of E ⊕ F, and since the mapping (e, Ae) ∈ ΓA 7→ e ∈ E is a continuous isomorphism, its inverse e ∈ E 7→ (e, Ae) ∈ ΓA is also continuous by Theorem 2.2.16. Since (e, Ae) ∈ ΓA 7→ Ae ∈ F is clearly continuous, so is the composition e 7→ (e, Ae) 7→ Ae.  The Closed graph theorem is often used in the following way. To show that a linear map A : E → F is continuous for E and F Banach spaces, it suffices to show that if en → 0 and Aen → e0 , then e0 = 0. 2.2.18 Corollary. Let E be a Banach space and F a closed subspace of E. Then F is split iff there exists P ∈ L(E, E) such that P ◦ P = P and F = { e ∈ E | P e = e }. Proof. If such a P exists, then clearly ker(P ) is a closed subspace of E that is an algebraic complement of F; any e ∈ E is of the form e = e − P e + P e with e − P e ∈ ker(P ) and P e ∈ F. Conversely, if E = F ⊕ G, define P : E → E by P (e) = e1 , where e = e1 + e2 , e1 ∈ F, e2 ∈ G. P is clearly linear, P 2 = P , and F = { e ∈ E | P e = e }, so all there is to show is that P is continuous. Let en = e1n + e2n → 0 and P (en ) = e1n → e0 ; that is, −e2n → e0 , and since F and G are closed this implies that e0 ∈ F ∩ G = {0}. By the closed graph theorem, P ∈ L(E, E).  2.2.19 Theorem (Fundamental Isomorphism Theorem). Suppose that the linear map A ∈ L(E, F) is surjective, where E and F are Banach spaces. Then E/ ker A and F are isomorphic Banach spaces. Proof.

The map [e] 7→ Ae is bijective and continuous (since its norm is ≤ kAk), so it is a homeomorphism. 

A sequence of maps A

Ai+1

· · · → Ei−1 →i Ei → Ei+1 → · · · of Banach spaces is said to be split exact if for all i, ker Ai+1 = range Ai and both ker Ai and range Ai split. With this terminology, Theorem 2.2.19 can be reformulated in the following way: If 0 → G → E → F → 0 is a split exact sequence of Banach spaces, then E/G is a Banach space isomorphic to F (thus F ∼ = G ⊕ F).

2.2 Linear and Multilinear Mappings

63

Uniform Boundedness Principle. Next we prove the uniform boundedness principle of Banach and Steinhaus, the third pillar of linear analysis. 2.2.20 Theorem. Let E and F be normed vector spaces, with E complete, and let {Ai }i∈I ⊂ L(E, F). If for each e ∈ E the set {kAi ek}i∈I is bounded in F, then {kAi k}i∈I is a bounded set of real numbers. Proof.

Let ϕ(e) = sup{ kAi ek | i ∈ I } and note that Sn = { e ∈ E | ϕ(e) ≤ n } =

\

{ e ∈ E | kAi ek ≤ n }

i∈I

S is closed and n≥1 Sn = E. Since E is a complete metric space, the Baire category theorem 1.7.4 says that some Sn has nonempty interior; that is, there exist r > 0 and e0 ∈ E such that ϕ(e) ≤ M, for all e ∈ cl(Dr (e0 )), where M > 0 is come constant. For each i ∈ I and kek = 1, we have kAi (re + e0 )k ≤ ϕ(re + e0 ) ≤ M , so that 1 1 1 kAi ek = kAi (re + e0 − e0 )k ≤ kAi (re + e0 )k + kAi e0 k r r r 1 ≤ (M + ϕ(e0 )), r that is, kAi k ≤ (M + ϕ(e0 ))/r for all i ∈ I.



2.2.21 Corollary. If {An } ⊂ L(E, F) is a strongly convergent sequence (i.e., limn→∞ An e = Ae exists for every e ∈ E), then A ∈ L(E, F). Proof. A is clearly a linear map. Since {An e} is convergent, it is a bounded set for each e ∈ E, so that by Theorem 2.2.20, {kAn k} is bounded by, say, M > 0. But then kAek = lim kAn ek ≤ lim sup kAn k kek ≤ M kek; n→∞

n→∞

that is, A ∈ L(E, F).



Exercises  2.2-1.

If E = Rn and F = Rm with the standard norms, and A : E → F is a linear map, show that

(i) kAk is the square root of the absolute value of the largest eigenvalue of AAT , where AT is the transpose of A, and (ii) if n, m ≥ 2, this norm does not come from an inner product. Hint: Use Exercise 2.1-1.  2.2-2. Let E = F = Rn with the standard norms and A, B ∈ L(E, F). Let hA, Bi = trace(AB T ). Show that this is an inner product on L(E, F).  2.2-3.

Show that the map L(E, F) × L(F, E) → R;

gives a (natural) isomorphism L(E, F)∗ ∼ = L(F, E).

(A, B) 7→ trace(AB)

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2. Banach Spaces and Differential Calculus

 2.2-4. Let E, F, G be Banach spaces and D ⊂ E a linear subspace. A linear map A : D → F is called closed if its graph ΓA := {(x, Ax) ∈ E × F | x ∈ D} is a closed subset of E × F. If A : D ⊂ E → F, and B : D ⊂ E → G are two closed operators with the same domain D, show that there are constants M1 , M2 > 0 such that kAek ≤ M1 (kBek + kek)

and kBek ≤ M2 (kAek + kek)

for all e ∈ E. Hint: Norm E ⊕ G by k(e, g)k = kek + kgk and define T : ΓB → G by T (e, Be) = Ae. Use the closed graph theorem to show that T ∈ L(ΓB , G).  2.2-5 (Linear transversality). Let E, F be Banach spaces, F0 ⊂ F a closed subspace, and T ∈ L(E, F). T is said to be transversal to F0 , if T −1 (F0 ) splits in E and T (E) + F0 = { T e + f | e ∈ E, f ∈ F } = F. Prove the following. (i) T is transversal to F0 iff π ◦ T ∈ L(E, F/F0 ) is surjective with split kernel; here π : F → F/F0 is the projection. (ii) If π ◦ T ∈ L(E, F/F0 ) is surjective and F0 has finite codimension, then ker(π ◦ T ) has the same codimension and T is transversal to F0 . ∼ (T (E) + F0 )/F0 to show E/ ker(π ◦ T ) ∼ Hint: Use the algebraic isomorphism T (E)/(F0 ∩ T (E)) = = F/F0 ; now use Corollary 2.2.18. (iii) If π ◦ T ∈ L(E, F/F0 ) is surjective and if ker T and F0 are finite dimensional, then ker(π ◦ T ) is finite dimensional and T is transversal to F0 . Hint: Use the exact sequence 0 → ker T → ker(π ◦ T ) → F0 ∩ T (E) → 0.  2.2-6.

Let E and F be Banach spaces. Prove the following.

(i) If f ∈ cl(S([a, b], L(E, F))) and e ∈ E, then b

Z

f (t)e dt = a

!

b

Z

f (t) dt (e) a

Hint: T 7→ T e is in L(L(E, F), F). (ii) If f ∈ cl(S([a, b], R) and v ∈ F, then Z

b

f (t)v dt = a

!

b

Z

f (t) dt (v) a

Hint: t 7→ multiplication by t in F is in L(R, L(F, F)); apply (i). (iii) Let X be a topological space and f : [a, b] × X → E be continuous. Then the mapping Z b g : X → E, g(x) = f (t, x)dt a

is continuous. Hint: For t ∈ R, x0 ∈ X and ε > 0 given, kf (s, x) − f (t, x0 )k < ε

if (s, x) ∈ Ut × Ux0 ,t ;

use compactness of [a, b] to find Ux0 as a finite intersection and such that kf (t, x) − f (t, x0 )k < ε for all t ∈ [a, b], x ∈ Ux0 .

2.2 Linear and Multilinear Mappings

65

 2.2-7. Show that the Banach isomorphism theorem is false for normed incomplete vector spaces in the following way. Let E be the space of all polynomials over R normed as follows: ka0 + a1 x + · · · + an xn k = max{|a0 |, . . . , |an |}. (i) Show that E is not complete. (ii) Define A : E → E by A

n X

! i

ai x

= a0 +

i=0

n X ai i=1

i

xi

and show that A ∈ L(E, E). Prove that A−1 : E → E exists. (iii) Show that A−1 is not continuous.  2.2-8. Let E and F be Banach spaces and A ∈ L(E, F). If A(E) has finite codimension, show that it is closed. Hint: If F0 is an algebraic complement to A(E) in F, show there is a continuous linear isomorphism E/ ker A ∼ = F/F0 ; compose its inverse with E/ ker A → A(E).  2.2-9 (Symmetrization operator).

Define Symk : Lk (E, F) → Lk (E, F),

by Symk A =

1 X σA, k! a∈Sk

where (σA)(e1 , . . . , ek ) = A(eσ(1) , . . . , eσ(k) ). Show that: (i) Symk (Lk (E, F)) = Lks (E, F). (ii) (Symk )2 = Symk . (iii) k Symk k ≤ 1. (iv) If F is Banach, then Lks (E, F) splits in Lk (E, F). Hint: Use Corollary 2.2.18. (v) (Symk A)0 = A0 .  2.2-10. Show that a k-multilinear map continuous in each argument separately is continuous. Hint: For k = 2: If ke1 k ≤ 1, then kA(e1 , e2 )k ≤ kA(·, e2 )k, which by the uniform boundedness principle implies the inequality kA(e1 , ·)k ≤ M for ke1 k ≤ 1.  2.2-11. (i) Prove the Mazur-Ulam Theorem following the steps below (see Mazur and Ulam [1932]; Banach [1932] p. 166] and V¨ ais¨ al¨ a [2003]): Every isometric surjective mapping ϕ : E → F such that ϕ(0) = 0 is a linear map. Here E and F are normed vector spaces; ϕ being isometric means that kϕ(x) − ϕ(y)k = kx − yk for all x, y ∈ E. (a) Fix x1 , x2 ∈ E and define   1 H1 = x kx − x1 k = kx − x2 k = kx1 − x2 k , 2   1 Hn = x ∈ Hn−1 kx − zk ≤ diam(Hn−1 ), z ∈ Hn−1 . 2

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2. Banach Spaces and Differential Calculus

Show that

1 1 diam(H1 ) ≤ n−1 kx1 − x2 k. 2n−1 2 T Conclude that if n≥1 Hn = 6 ∅, then it consists of one point only. diam(Hn ) ≤

(b) Show by induction that if x ∈ Hn , then x1 + x2 − x ∈ Hn . T (c) Show that n≥1 Hn = {(x1 + x2 )/2}. Hint: Show inductively that (x1 + x2 )/2 ∈ Hn using (b). (d) From (c) deduce that  ϕ

 1 1 (x1 + x2 ) = (ϕ(x1 ) + ϕ(x2 )). 2 2

Use ϕ(0) = 0 to conclude that ϕ is linear. (ii) (Chernoff, 1970). The goal of this exercise is to study the Mazur–Ulam theorem, dropping the assumption that ϕ is onto, and replacing it with the assumption that ϕ is homogeneous: ϕ(tx) = tϕ(x) for all t ∈ R and x ∈ E. (a) A normed vector space is called strictly convex if equality holds in the triangle inequality only for colinear points. Show that if F is strictly convex, then ϕ is linear. Hint:

   

x+y x+y



kϕ(x) − ϕ(y)k = ϕ(x) − ϕ

+ ϕ(y) − ϕ

2 2 and

   



ϕ(x) − ϕ x + y = ϕ(y) − ϕ x + y .



2 2

Show that

 ϕ

x+y 2

 =

1 (ϕ(x) + ϕ(y)). 2

(b) Show that, in general, the assumption on ϕ being onto is necessary by considering the following counterexample. Let E = R2 and F = R3 , both with the max norm. Define ϕ : E → F by √ ϕ(a, b) = (a, b, ab), a, b > 0; √ a, b > 0; ϕ(−a, b) = (−a, b, − ab), √ ϕ(a, −b) = (a, −b, − ab), a, b > 0; √ ϕ(−a, b) = (−a, −b, − ab), a, b > 0. Show that ϕ is not linear, ϕ is homogeneous, ϕ is an isometry, and ϕ(0, 0) = (0, 0, 0). Hint: Prove the inequality |αβ − γδ| ≤ max(|α2 − γ 2 |, |β 2 − δ 2 |).  2.2-12.

Let E be a complex n-dimensional vector space.

(i) Show that the set of all operators A ∈ L(E, E) which have n distinct eigenvalues is open and dense in E. Hint: Let p be the characteristic polynomial of A, that is, p(λ) = det (A − λI), and let µ1 , . . . , µn−1 be the roots of p0 . Then A has multiple eigenvalues iff p(µ1 ) · · · p(µn−1 ) = 0. The last expression is a symmetric polynomial in µ1 , . . . , µn−1 , and so is a polynomial in the coefficients of p0 and therefore is a polynomial q in the entries of the matrix of A in a basis. Show that q −1 (0) is the set of complex n × n matrices which have multiple eigenvalues; q −1 (0) has open dense complement by Exercise 1.1-12.

2.2 Linear and Multilinear Mappings

67

(ii) Prove the Cayley–Hamilton Theorem: If p is the characteristic polynomial of A ∈ L(E, E), then p(A) = 0 . Hint: If the eigenvalues of A are distinct, show that the matrix of A in the basis of eigenvectors e1 , . . . , en is diagonal. Apply A, A2 ,. . . , An−1 . Then show that for any polynomial q the matrix of q(A) in the same basis is diagonal with entries q(λi ), where λi are the eigenvalues of A. Finally, let q = p. If A is general, apply (i).  2.2-13.

Let E be a normed real (resp. complex) vector space.

(i) Show that λ : E → R (resp., C) is continuous if and only if ker λ is closed. Hint: Let e ∈ E satisfy λ(e) = 1 and choose a disk D of radius r centered at e such that D∩(e+ker λ) = ∅. Then λ(x) 6= 1 for all x ∈ D. Show that if x ∈ D then λ(x) < 1. If not, let α = λ(x), |α| > 1. Then kx/αk < r and λ(x/α) = 1. (ii) Show that if F is a closed subspace of E and G is a finite dimensional subspace, then G + F is closed. Hint: Assume G is one dimensional and generated by g. Write any x ∈ G + F as x = λ(x)g + f and use (i) to show λ is continuous on G + F.  2.2-14.

Let F be a Banach space.

(i) Show that if E is a finite dimensional subspace of F, then E is split. Hint: Define P : F → F by P (x) =

X

ei (x)ei ,

i=1,...,n

where {e1 , . . . , en } is a basis of E and {e1 , . . . , en } is a dual basis, that is, ei (ej ) = δij . Then use Corollary 2.2.18. (ii) Show that if E is closed and finite codimensional, then it is split. (iii) Show that if E is closed and contains a finite-codimensional subspace G of F, then it is split. (iv) Let λ : F → R be a linear discontinuous map and let E = ker λ. Show that the codimension of E is 1 and that E is not closed. Thus finite codimensional subspaces of F are not necessarily closed. Compare this with (i) and (ii), and with Exercise 2.2-8.  2.2-15. Let E and F be Banach spaces and T ∈ L(E, F). Define T ∗ : F∗ → E∗ by hT ∗ β, ei = hβ, T ei for e ∈ E, β ∈ F∗ . Show that: (i) T ∗ ∈ L(F∗ , E∗ ) and T ∗∗ |E = T . (ii) ker T ∗ = T (E)o := { β ∈ F∗ | hβ, T ei = 0 for all e ∈ E } and ker T = (T ∗ (F∗ ))o := { e ∈ E | hT ∗ β, ei = 0 for all β ∈ F∗ }. (iii) If T (E) is closed, then T ∗ (F∗ ) = (ker T )o . Hint: The induced map E/ ker T → T (E) is a Banach space isomorphism; let S be its inverse. If λ ∈ (ker T )o , define the element µ ∈ (E/ ker T )∗ by µ([e]) = λ(e). Let ν ∈ F∗ denote the extension of S ∗ (µ) ∈ (T (E))∗ to ν ∈ F∗ with the same norm and show that T ∗ (ν) = λ. (iv) If T (E) is closed, then ker T ∗ is isomorphic to (F/T (E))∗ and (ker T )∗ is isomorphic to E∗ /T ∗ (F∗ ).

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2. Banach Spaces and Differential Calculus

2.3

The Derivative

Definition of the Derivative. For a differentiable function f : U ⊂ R → R, the usual interpretation of the derivative at a point u0 ∈ U is the slope of the line tangent to the graph of f at u0 . To generalize this, we interpret Df (u0 ) = f 0 (u0 ) as a linear map acting on the vector (u − u0 ). 2.3.1 Definition. Let E, F be normed vector spaces, U be an open subset of E and let f : U ⊂ E → F be a given mapping. Let u0 ∈ U . We say that f is differentiable at the point u0 provided there is a bounded linear map Df (u0 ) : E → F such that for every  > 0, there is a δ > 0 such that whenever 0 < ku − u0 k < δ, we have kf (u) − f (u0 ) − Df (u0 ) · (u − u0 )k < , ku − u0 k where k · k represents the norm on the appropriate space and where the evaluation of Df (u0 ) on e ∈ E is denoted Df (u0 ) · e. This definition can also be written as lim

u→u0

f (u) − f (u0 ) − Df (u0 ) · (u − u0 ) = 0. ku − u0 k

We shall shortly show that the derivative is unique if it exists and embark on relating this notion to ones that are perhaps more familiar to the reader in Euclidean space; we shall also develop many familiar properties of the derivative. However, it is useful to first slightly rephrase the definition. We shall do this in terms of the notion of tangency. Tangency of Maps. An alternative way to think of the derivative in one variable calculus is to say that Df (u0 ) is the unique linear map from R into R such that the mapping g : U → R given by u 7→ g(u) = f (u0 ) + Df (u0 ) · (u − u0 ) is tangent to f at u0 , as in Figure 2.3.1. g

f

U

Figure 2.3.1. Derivative of a function of one variable

2.3.2 Definition. Let E, F be normed vector spaces, with maps f, g : U ⊂ E → F where U is open in E. We say f and g are tangent at the point u0 ∈ U if f (u0 ) = g(u0 ) and lim

u→u0

kf (u) − g(u)k = 0, ku − u0 k

where k · k represents the norm on the appropriate space.

2.3 The Derivative

69

2.3.3 Proposition. For f : U ⊂ E → F and u0 ∈ U there is at most one L ∈ L(E, F) such that the map gL : U ⊂ E → F given by gL (u) = f (u0 ) + L(u − u0 ) is tangent to f at u0 . Proof. Let L1 and L2 ∈ L(E, F) satisfy the conditions of the proposition. Suppose that e ∈ E is a unit vector so that kek = 1. Let u = u0 + λe for λ ∈ R (or C). Then for λ 6= 0, with |λ| small enough so that u ∈ U , we have kL1 (u − u0 ) − L2 (u − u0 )k ku − u0 k kf (u) − f (u0 ) − L1 (u − u0 )k ≤ ku − u0 k kf (u) − f (u0 ) − L2 (u − u0 )k + . ku − u0 k

kL1 e − L2 ek =

As λ → 0, the right hand side approaches zero and therefore k(L1 − L2 )ek = 0 for all e ∈ E satisfying kek = 1; therefore, kL1 − L2 k = 0 and thus L1 = L2 .  We can thus rephrase the definition of the derivative this way: If, in Proposition 2.3.3, there is such an L ∈ L(E, F), then f is differentiable at u0 , and the derivative of f at u0 is Df (u0 ) = L. Thus, the derivative, if it exists, is unique. 2.3.4 Definition.

If f is differentiable at each u0 ∈ U , the map Df : U → L(E, F);

u 7→ Df (u)

is called the derivative of f . Moreover, if Df is a continuous map (where L(E, F) has the norm topology), we say f is of class C 1 (or is continuously differentiable). Proceeding inductively we define Dr f := D(Dr−1 f ) : U ⊂ E → Lr (E, F) if it exists, where we have identified L(E, Lr−1 (E, F)) with Lr (E, F) (see Proposition 2.2.9). If Dr f exists and is norm continuous, we say f is of class C r . Basic Properties of the Derivative. We shall reformulate the definition of the derivative with the aid of the somewhat imprecise but very convenient Landau symbol : o(ek ) will denote a continuous function of e defined in a neighborhood of the origin of a normed vector space E, satisfying lime→0 (o(ek )/kekk ) = 0. The collection of these functions forms a vector space. Clearly f : U ⊂ E → F is differentiable at u0 ∈ U iff there exists a linear map Df (u0 ) ∈ L(E, F) such that f (u0 + e) = f (u0 ) + Df (u0 ) · e + o(e). Let us use this notation to show that if Df (u0 ) exists, then f is continuous at u0 : lim f (u0 + e) = lim (f (u0 ) + Df (u0 ) · e + o(e)) = f (u0 ).

e→0

e→0

2.3.5 Proposition (Linearity of the Derivative). Let f, g : U ⊂ E → F be r times differentiable mappings and a a real (or complex ) constant. Then af and f + g : U ⊂ E → F are r times differentiable with Dr (f + g) = Dr f + Dr g Proof.

and

Dr (af ) = aDr f.

If u ∈ U and e ∈ E, then f (u + e) = f (u) + Df (u) · e + o(e) and g(u + e) = g(u) + Dg(u) · e + o(e),

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2. Banach Spaces and Differential Calculus

so that adding these two relations yields (f + g)(u + e) = (f + g)(u) + (Df (u) + Dg(u)) · e + o(e). The case r > 1 follows by induction. Similarly, af (u + e) = af (u) + aDf (u) · e + ao(e) = af (u) + aDf (u) · e + o(e).  2.3.6 Proposition (Derivative of a Cartesian Product). Let fi : U ⊂ E → Fi , 1 ≤ i ≤ n, be a collection of r times differentiable mappings. Then f = f1 × · · · × fn : U ⊂ E → F1 × · · · × Fn defined by f (u) = (f1 (u), . . . , fn (u)) is r times differentiable and Dr f = Dr f1 × · · · × Dr fn . Proof.

For u ∈ U and e ∈ E, we have f (u + e) = (f1 (u + e), . . . , fn (u + e)) = (f1 (u) + Df1 (u) · e + o(e), . . . , fn (u) + Dfn (u) · e + o(e)) = (f1 (u), . . . , fn (u)) + (Df1 (u), . . . , Dfn (u)) · e +(o(e), . . . , o(e)) = f (u) + Df (u) · e + o(e),

the last equality follows using the sum norm in F1 × · · · × Fn : k(o(e), . . . , o(e))k = ko(e)k + · · · + ko(e)k, so (o(e), . . . , o(e)) = o(e).



Notice from the definition that for L ∈ L(E, F), DL(u) = L for any u ∈ E. It is also clear that the derivative of a constant map is zero. Usually all our spaces will be real and linearity will mean real-linearity. In the complex case, differentiable mappings are the subject of analytic function theory, a subject we shall not pursue in this book (see Exercise 2.3-6 for a hint of why there is a relationship with analytic function theory). Jacobian Matrices. In addition to the foregoing approach, there is a more traditional way to differentiate a function f : U ⊂ Rn → Rm . We write out f in component form using the following notation: f (x1 , . . . , xn ) = (f 1 (x1 , . . . , xn ), . . . , f m (x1 , . . . , xn )) and compute partial derivatives, ∂f j /∂xi for j = 1, . . . , m and i = 1, . . . , n, where the symbol ∂f j /∂xi means that we compute the usual derivative of f j with respect to xi while keeping the other variables x1 , . . . , xi−1 , xi+1 , . . . , xn fixed. For f : R → R, Df (x) is just the linear map “multiplication by df /dx,” that is, df /dx = Df (x) · 1. This fact, which is obvious from the definitions, can be generalized to the following statement. 2.3.7 Proposition. Suppose that U ⊂ Rn is an open set and that f : U → Rm is differentiable. Then the partial derivatives ∂f j /∂xi exist. Moreover, the matrix of the linear map Df (x) with respect to the standard

2.3 The Derivative

bases in Rn and Rm is given by

∂f 1  ∂x1   ∂f 2   1  ∂x   .  ..    m ∂f ∂x1 where each partial derivative is evaluated at x of f . 

∂f 1 ∂x2 ∂f 2 ∂x2 .. .

71

 ∂f 1 ∂xn   ∂f 2   ···  ∂xn  , ..  .    m m ∂f ∂f ··· ∂x2 ∂xn = (x1 , . . . , xn ). This matrix is called the Jacobian matrix ···

Proof. By the usual definition of the matrix of a linear mapping from linear algebra, the (j, i)th matrix element aji of Df (x) is given by the jth component of the vector Df (x) · ei , where e1 , . . . , en is the standard basis of Rn . Letting y = x + hei , we see that kf (y) − f (x) − Df (x) · (y − x)k ky − xk

1

f (x1 , . . . , xi + h, . . . , xn ) − f (x1 , . . . , xn ) − hDf (x) · ei = |h| approaches zero as h → 0, so the jth component of the numerator does as well; that is, 1 j 1 j lim f (x , . . . , xi + h, . . . , xn ) − f j (x1 , . . . , xn ) − hai = 0, h→0 |h| which means that aji = ∂f j /∂xi .



In computations one can usually compute the Jacobian matrix easily, and this proposition then gives Df . In some books, Df is called the differential or the total derivative of f . 2.3.8 Example. Let f : R2 → R3 , f (x, y) = (x2 , x3 y, x4 y 2 ). Then Df (x, y) is the linear map whose matrix in the standard basis is   1 ∂f 1 ∂f    ∂x ∂y  2x 0     2  ∂f 2    ∂f 3x2 y x3  =  ,    ∂x ∂y   3 2 4 4x y 2x y  ∂f 3 ∂f 3  ∂x 1

2

2

3

∂y

3

where f (x, y) = x , f (x, y) = x y, f (x, y) = x4 y 2 .



One should take special note when m = 1, in which case we have a real–valued function of n variables. Then Df has the matrix   ∂f ∂f ··· n ∂x1 ∂x and the derivative applied to a vector e = (a1 , . . . , an ) is Df (x) · e =

n X ∂f i a, i ∂x i=1

where it is understood that the partial derivatives are evaluated at x.

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2. Banach Spaces and Differential Calculus

The Gradient and Differential. It should be emphasized that Df assigns a linear mapping to each x ∈ U and the definition of Df (x) is independent of the basis used. If we change the basis from the standard basis to another one, the matrix elements will of course change. If one examines the definition of the matrix of a linear transformation, it can be seen that the columns of the matrix relative to the new basis will be the derivative Df (x) applied to the new basis in Rn with this image vector expressed in the new basis in Rm . Of course, the linear map Df (x) itself does not change from basis to basis. In the case m = 1, Df (x) is, in the standard basis, a 1 × n matrix. The vector whose components are the same as those of Df (x) is called the gradient of f , and is denoted grad f or ∇f . Thus for f : U ⊂ Rn → R,   ∂f ∂f ,··· , n grad f = ∂x1 ∂x (Sometimes it is said that grad f is just Df with commas inserted!) The formation of gradients makes sense in a general inner product space as follows. 2.3.9 Definition. (i) Let E be a normed space and f : U ⊂ E → R be differentiable. Thus, Df (u) ∈ L(E, R) = E∗ . In this case we sometimes write df (u) for Df (u) and call df the differential of f . Thus df : U → E∗ . (ii) If E is a Hilbert space, the gradient of f is the map grad f = ∇f : U → E

defined by h∇f (u), ei = df (u) · e,

where df (u) · e means the linear map df (u) applied to the vector e. Note that the existence of ∇f (u) requires the Riesz representation theorem (see Theorem 2.2.5). The notation δf /δu instead of (grad f )(u) = ∇f (u) is also in wide use, especially in the case in which E is a space of functions. See Supplement 2.4C below. 2.3.10 Example.

Let (E, h, i) be a real inner product space and let f (u) = kuk2 . Since kuk2 = ku0 k2 + 2 hu0 , u − u0 i + ku − u0 k2 ,

we obtain df (u0 ) · e = 2 hu0 , ei and thus ∇f (u) = 2u. Hence f is of class C 1 . But since Df (u) = 2 hu, ·i ∈ E∗ is a continuous linear map in u ∈ E, it follows that D2 f (u) = Df ∈ L(E, E∗ ) and thus Dk f = 0 for k ≥ 3. Thus f is of class C ∞ . The mapping f considered here is a special case of a polynomial mapping (see Definition 2.2.10).  Fundamental Theorem. We close this section with the fundamental theorem of calculus in the context of real Banach spaces. First a bit of notation. If ϕ : U ⊂ R → F is differentiable, then Dϕ(t) ∈ L(R, F). However, the space L(R, F) is isomorphic to F via the isomorphism A 7→ A(1). Note that kAk = kA(1)k, so the isomorphism preserves the norm. We denote dϕ = Dϕ(t) · 1, 1 ∈ R dt ϕ(t + h) − ϕ(t) ϕ0 (t) = lim h→0 h ϕ0 =

and ϕ is differentiable iff ϕ0 exists. 2.3.11 Theorem (Fundamental Theorem of Calculus). (i) If g : [a, b] → F is continuous, where F is a real normed space, then the map Z t f : ]a, b[ → F defined by f (t) = g(s) ds a 0

is differentiable and f = g.

2.3 The Derivative

73

(ii) If f : [a, b] → F is continuous, is differentiable on the open interval ]a, b[ and if f 0 extends to a continuous map on [a, b], then Z b f (b) − f (a) = f 0 (s) ds. a

Proof.

We first prove part (i). Let t0 ∈ ]a, b[. Since the integral is linear and continuous,

Z

t0 +h

(g(s) − g(t0 ))ds ≤ |h|Lg,h , kf (t0 + h) − f (t0 ) − hg(t0 )k =

t0

where Lg,h = sup{ kg(s) − g(t0 )k | t0 ≤ s ≤ t0 + h }. However, Lg,h → 0 as |h| → 0 by continuity of g at t0 . Turning now to part (ii), let the function h(t) be defined by Z t  0 f (s) ds − f (t), h(t) = a

By (i), h0 (t) = 0 on ]a, b[ and h is continuous on [a, b]. If for some t ∈ [a, b], h(t) 6= h(a), then by the Hahn–Banach theorem there exists α ∈ F∗ such that (α ◦ h)(t) 6= (α ◦ h)(a). Moreover, α ◦ h is differentiable on ]a, b[ and its derivative is zero (Exercise 2.3-4). Thus by elementary calculus, α ◦ h is constant on [a, b], a contradiction. Hence h(t) = h(a) for all t ∈ [a, b]. In particular, h(a) = h(b) 

Exercises  2.3-1.

Let B : E × F → G be a continuous bilinear map of normed spaces. Show that B is C ∞ and that DB(u, v)(e, f ) = B(u, f ) + B(e, v).

 2.3-2.

Show that the derivative of a map is unaltered if the spaces are renormed with equivalent norms.

 2.3-3.

If f ∈ S k (E, F), show that for, i = 1, . . . , k, ∂k D f (0)(e1 , . . . , ek ) = f (t1 e1 + · · · + tk ek ) ∂t1 · · · ∂tk t1 =···=tk =0 k

and Di f (0) = 0

for i = 1, . . . , k − 1.

 2.3-4. Let f : U ⊂ E → F be a differentiable (resp., C r ) map and A ∈ L(F, G). Show that A ◦ f : U ⊂ E → G is differentiable (resp., C r ) and Dr (A ◦ f )(u) = A ◦ Dr f (u). Hint: Use induction.  2.3-5.

Let f : U ⊂ E → F be r times differentiable and A ∈ L(G, E). Show that Di (f ◦ A)(v) · (g1 , . . . , gi ) = Di f (Av) · (Ag1 , . . . , Agi )

exists for all i ≤ r, where v ∈ A−1 (U ), and g1 , . . . , gi ∈ G. Generalize to the case where A is an affine map.  2.3-6. (i) Show that a complex linear map A ∈ L(C, C) is necessarily of the form A(z) = λz, for some λ ∈ C. (ii) Show that the matrix of A ∈ L(C, C), when A is regarded as a real linear map in L(R2 , R2 ), is of the form   a −b . b a Hint: λ = a + ib.

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2. Banach Spaces and Differential Calculus

(iii) Show that a map f : U ⊂ C → C, f = g + ih, g, h : U ⊂ R2 → R is complex differentiable iff the Cauchy–Riemann equations ∂g ∂h ∂g ∂h = , =− ∂x ∂y ∂y ∂x are satisfied. Hint: Use (ii) and Proposition 2.3.7.  2.3-7. Let (E, h, i) be a complex inner product space. Show that the map f (u) = kuk2 is not differentiable. Contrast this with Example 2.3.10. Hint: Df (u), if it exists, should equal 2 Re(hu, ·i).  2.3-8.

Show that the matrix of D2 f (x) ∈ L2 (Rn , R) for f : U ⊂ Rn → R, is given by ∂2f  ∂x1 ∂x1  ..   .   2 ∂ f ∂xn ∂x1 

∂2f ∂x1 ∂x2 .. . ∂2f ∂xn ∂x2

···

···

 ∂2f ∂x1 ∂xn   ..  . .   ∂2f ∂xn ∂xn

Hint: Apply Proposition 2.3.7. Recall that the matrix of a bilinear mapping B ∈ L(Rn , Rm ; R) has the entries B(ei , fj ) (first index = row index, second index = column index), where {e1 , . . . , en } and {f1 , . . . , fm } are ordered bases of Rn and Rm , respectively.

2.4

Properties of the Derivative

In this section some of the fundamental properties of the derivative are developed. These properties are analogues of rules familiar from elementary calculus. Differentiability implies Lipschitz. continuity.

Let us begin by strengthening the fact that differentiability implies

2.4.1 Proposition. Suppose U ⊂ E is open and f : U → F is differentiable on U . Then f is continuous. In fact, for each u0 ∈ U there is a constant M > 0 and a δ0 > 0 with the property that ku − u0 k < δ0 implies kf (u) − f (u0 )k ≤ M ku − u0 k. (This is called the Lipschitz property .) Proof.

Using the general inequality |ke1 k − ke2 k| ≤ ke1 − e2 k, we get | kf (u) − f (u0 )k − kDf (u0 ) · (u − u0 )k | ≤ kf (u) − f (u0 ) − Df (u0 ) · (u − u0 )k = ko(u − u0 )k ≤ ku − u0 k

for ku − u0 k ≤ δ0 , where δ0 is some positive constant depending on u0 ; this holds since lim

u→u0

o(u − u0 ) = 0. ku − u0 k

Thus, kf (u) − f (u0 )k ≤ kDf (u0 ) · (u − u0 )k + ku − u0 k ≤ (kDf (u0 )k + 1)ku − u0 k for ku − u0 k ≤ δ0 .



2.4 Properties of the Derivative

75

Chain Rule. Perhaps the most important rule of differential calculus is the chain rule. To facilitate its statement, the notion of the tangent of a map is introduced. The text will begin conceptually distinguishing points in U from vectors in E. At this point it is not so clear that the distinction is important, but it will help with the transition to manifolds in Chapter 3. 2.4.2 Definition.

Suppose f : U ⊂ E → F is of class C 1 . Define the tangent of f to be the map Tf : U × E → F × F

given by T f (u, e) = (f (u), Df (u) · e),

where we recall that Df (u) · e denotes Df (u) applied to e ∈ E as a linear map. If f is of class C r , define T r f = T (T r−1 f ) inductively. From a geometric point of view, T f is a more “natural” object than D. The reasons for this will become clearer as we proceed, but roughly speaking, the essence is this: if we think of (u, e) as a vector with base point u and vector part e, then (f (u), Df (u) · e) is the image vector with its base point f (u), as in Figure 2.4.1. Another reason for this is the simple and elegant behavior of T under composition, as given in the next theorem.

e E u

f

F

Figure 2.4.1. The geometry of the tangent map

2.4.3 Theorem (C r Composite Mapping Theorem). Suppose f : U ⊂ E → V ⊂ F and g : V ⊂ F → G are differentiable (resp., C r ) maps. Then the composite g ◦ f : U ⊂ E → G is also differentiable (resp., C r ) and T (g ◦ f ) = T g ◦ T f, (resp., T r (g ◦ f ) = T r g ◦ T r f ). The formula T (g ◦ f ) = T g ◦ T f is equivalent to the following equality in terms of the usual derivative D: D(g ◦ f )(u) = Dg(f (u)) ◦ Df (u). This statement or the statement in terms of T , namely T (g ◦ f ) = T g ◦ T f is called the chain rule. Proof.

Since f is differentiable at u ∈ U and g is differentiable at f (u) ∈ V , we have f (u + e) = f (u) + Df (u) · e + o(e)

for e ∈ E

and for v = f (u) and w ∈ F , we have g(v + w) = g(v) + Dg(v) · w + o(w). Thus, (g ◦ f )(u + e) = g(f (u) + Df (u) · e + o(e)) = (g ◦ f )(u) + Dg(f (u)) · (Df (u) · e) +Dg(f (u)) · (o(e)) + o(Df (u) · e + o(e)).

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2. Banach Spaces and Differential Calculus

For e in a neighborhood of the origin, kDf (u) · e + o(e)k ≤ kek

 kDf (u)k +

ko(e)k kek

 ≤M

for some constant M > 0, and kDg(f (u)) · o(e)k ≤ kDg(f (u))k ko(e)k. Therefore, k(o(Df (u) · e + o(e)))k kDf (u) · e + o(e)k ko(Df (u) · e + o(e))k = · kek kDf (u) · e + o(e)k kek k(o(Df (u) · e + o(e)))k ≤M . kDf (u) · e + o(e)k Hence, we conclude that Dg(f (u)) · (o(e)) + o(Df (u) · e + o(e)) = o(e) and thus D(g ◦ f )(u) · e = Dg(f (u)) · (Df (u) · e). Denote by ϕ : L(F, G) × L(E, F) → L(E, G) the bilinear mapping ϕ(B, A) = B ◦ A and note that ϕ ∈ L(L(F, G), L(E, F); L(E, G)) since kB ◦ Ak ≤ kBk kAk; that is, kϕk ≤ 1. Let (Dg ◦ f ) × Df : U → L(F, G) × L(E, F) be defined by [(Dg ◦ f ) × Df ](u) = (Dg(f (u)), Df (u)); notice that this map is continuous if f and g are of class C 1 . Therefore the composite function ϕ ◦ ((Dg ◦ f ) × Df ) = D(g ◦ f ) : U → L(E, G) is continuous if f and g are C 1 , that is, g ◦ f is C 1 . Inductively suppose f and g are C r . Then Dg is C r−1 , so Dg ◦ f is C r−1 and thus the map (Dg ◦ f ) × Df is C r−1 (see Proposition 2.3.6). Since ϕ is C ∞ (Exercise 2.3-1), again the inductive hypothesis forces ϕ ◦ ((Dg ◦ f ) × Df ) = D(g ◦ f ) to be C r−1 ; that is, g ◦ f is C r . The formula T r (g ◦ f ) = T r g ◦ T r f is a direct verification for r = 1 using the chain rule, and the rest follows by induction.  If E = Rm , F = Rn , G = Rp , and f = (f 1 , . . . , f n ), g = (g 1 , . . . , g p ), where f i : U → R and g j : V → R, by Proposition 2.3.7 the chain rule becomes ∂(g ◦ f )1 (x) ···  ∂x1  ..   .   ∂(g ◦ f )p (x) ··· ∂x1 1 ∂g (f (x))  ∂y 1   .. = .   p  ∂g (f (x)) ∂y 1

 ∂(g ◦ f )1 (x)  ∂xm  ..   .   ∂(g ◦ f )p (x) ∂xm   1  ∂g 1 (f (x)) ∂f 1 (x) ∂f (x) ··· · · · ∂y n  ∂x1 ∂xm       .. . ..  · ..   . .       ∂g p (f (x))  ∂f n (x) ∂f n (x) ··· · · · ∂y n ∂x1 ∂xm which, when read componentwise, becomes the usual chain rule from calculus: 

n

∂(g ◦ f )j (x) X ∂g j (f (x)) ∂f k (x) = , ∂xi ∂y k ∂xi k=1

i = 1, . . . , m.

2.4 Properties of the Derivative

Product Rule. following.

77

The chain rule applied to B ∈ L(F1 , F2 ; G) and f1 × f2 : U ⊂ E → F1 × F2 yields the

2.4.4 Theorem (The Leibniz or Product Rule). Let fi : U ⊂ E → Fi , i = 1, 2, be differentiable (resp., C r ) maps and B ∈ L(F1 , F2 ; G). Then the mapping B(f1 , f2 ) = B ◦ (f1 × f2 ) : U ⊂ E → G is differentiable (resp., C r ) and D(B(f1 , f2 ))(u) · e = B(Df1 (u) · e, f2 (u)) + B(f1 (u), Df2 (u) · e). In the case F1 = F2 = R and B is multiplication, Theorem 2.4.4 reduces to the usual product rule for derivatives. Leibniz’ rule can easily be extended to multilinear mappings (Exercise 2.4-3). Directional Derivatives. derivative. 2.4.5 Definition. e ∈ E at u if

The first of several consequences of the chain rule involves the directional

Let f : U ⊂ E → F and let u ∈ U . We say that f has a derivative in the direction d f (u + te) dt t=0

exists. We call this element of F the directional derivative of f in the direction e at u. Sometimes a function all of whose directional derivatives exist is called Gˆ a teaux differentiable, whereas a function differentiable in the sense we have defined is called Fr´ e chet differentiable. The latter is stronger, according to the following. (See also Exercise 2.4-10.) 2.4.6 Proposition. by

If f is differentiable at u, then the directional derivatives of f exist at u and are given d f (u + te) = Df (u) · e. dt t=0

Proof. A path in E is a map from I into E, where I is an open interval of R. Thus, if c is differentiable, for t ∈ I we have Dc(t) ∈ L(R, E), by definition. Recall that we identify L(R, E) with E by associating Dc(t) with Dc(t) · 1 (1 ∈ R). Let dc (t) = Dc(t) · 1. dt For f : U ⊂ E → F of class C 1 we consider f ◦ c, where c : I → U . It follows from the chain rule that d dc (f (c(t))) = D(f ◦ c)(t) · 1 = Df (c(t)) · dt dt The proposition follows by choosing c(t) = u + te, where u, e ∈ E, I = ]−λ, λ[, and λ is sufficiently small.  For f : U ⊂ Rn → R, the directional derivative is given in terms of the standard basis {e1 , . . . , en } by Df (u) · e =

∂f 1 ∂f x + · · · + n xn , ∂x1 ∂x

where e = x1 e1 + · · · + xn en . This follows from Proposition 2.3.7 and Proposition 2.4.6. The formula in Proposition 2.4.6 is sometimes a convenient method for computing Df (u) · e. For example, let us compute the differential of a homogeneous polynomial of degree 2 from E to F. Let f (e) = A(e, e), where A ∈ L2 (E; F). By the chain and Leibniz rules, d Df (u) · e = A(u + te, u + te) = A(u, e) + A(e, u). dt t=0 If A is symmetric, then Df (u) · e = 2A(u, e).

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2. Banach Spaces and Differential Calculus

Mean Value Inequality.

One of the basic tools for finding estimates is the following.

2.4.7 Proposition. Let E and F be real Banach spaces, f : U ⊂ E → F a C 1 -map, x, y ∈ U , and c a C 1 arc in U connecting x to y; that is, c is a continuous map c : [0, 1] → U , which is C 1 on ]0, 1[, c(0) = x, and c(1) = y. Then Z 1 f (y) − f (x) = Df (c(t)) · c0 (t) dt. 0

If U is convex and c(t) = (1 − t)x + ty, then Z

1

Df ((1 − t)x + ty) · (y − x) dt  Z 1 Df ((1 − t)x + ty) dt · (y − x). =

f (y) − f (x) =

0

0

Proof. If g(t) = (f ◦ c)(t), the chain rule implies g 0 (t) = Df (c(t)) · c0 (t) and the fundamental theorem of calculus gives Z 1 g(1) − g(0) = g 0 (t) dt, 0

which is the first equality. The second equality for U convex and c(t) = (1 − t)x + ty is Exercise 2.2-6(i).  2.4.8 Proposition (Mean Value Inequality). Suppose U ⊂ E is convex and f : U ⊂ E → F is C 1 . Then for all x, y ∈ U   kf (y) − f (x)k ≤ sup kDf ((1 − t)x + ty)k ky − xk. 0≤t≤1

Thus, if kDf (u)k is uniformly bounded on U by a constant M > 0, then for all x, y ∈ U kf (y) − f (x)k ≤ M ky − xk. If F = R, then f (y) − f (x) = Df (c) · (y − x) for some c on the line joining x to y. Proof. The inequality follows directly from Proposition 2.4.7. The last assertion follows from the intermediate value theorem as in elementary calculus.  2.4.9 Corollary.

Let U ⊂ E be an open set; then the following are equivalent:

(i) U is connected ; (ii) every differentiable map f : U ⊂ E → F satisfying Df = 0 on U is constant. Proof.

If U = U1 ∪ U2 and U1 ∩ U2 = ∅, where U1 and U2 are open, then the mapping ( 0, if u ∈ U1 ; f (u) = e, if u ∈ U2 ,

where e ∈ F, e 6= 0 is a fixed vector, has Df = 0, yet is not constant. Conversely, assume that U is connected and Df = 0. Then f is in fact C ∞ . Let u0 ∈ U be fixed and consider the set S = { u ∈ U | f (u) = f (u0 ) }. Then S 6= ∅ (since u0 ∈ S), S ⊂ U , and S is closed since f is continuous. We shall show that S is also open. If u ∈ S, consider v ∈ Dr (u) ⊂ U and apply Proposition 2.4.8 to get kf (u) − f (v)k ≤ sup{ kDf ((1 − t)u + tv)k | t ∈ [0, 1] } ku − vk = 0; that is, f (v) = f (u) = f (u0 ) and hence Dr (u) ⊂ S. Connectedness of U implies S = U .



2.4 Properties of the Derivative

79

If f is Gˆateaux differentiable and the Gˆ ateaux derivative is in L(E, F); that is, for each u ∈ V there exists Gu ∈ L(E, F) such that d = Gu e, f (u + te) dt t=0 and if u 7→ Gu is continuous, we say f is C 1 -Gˆ ateaux . The mean value inequality holds, replacing C 1 1 everywhere by “C -Gˆ ateaux” and the identical proofs work. When studying differentiability the following is often useful. 2.4.10 Corollary. Proof.

If f : U ⊂ E → F is C 1 -Gˆ ateaux then it is C 1 and the two derivatives coincide.

Let u ∈ U and work in a disk centered at u. Proposition 2.4.7 gives

Z 1 

(G − G ) dt e kf (u + e) − f (u) − Gu ek = u+te u

0

≤ sup{ kGu+te − Gu k | t ∈ [0, 1] } kek and the sup converges to zero as, e → 0, by uniform continuity of the map t ∈ [0, 1] 7→ Gu+te ∈ L(E, F). This says that Df (u) · e exists and equals Gu e.  Partial Derivatives. being obvious.

We shall discuss only functions of two variables, the generalization to n variables

2.4.11 Definition. Let f : U → F be a mapping defined on the open set U ⊂ E1 ⊕ E2 and let u0 = (u01 , u02 ) ∈ U . The derivatives of the mappings v1 7→ f (v1 , u02 ), v2 7→ f (u01 , v2 ), where v1 ∈ E1 and v2 ∈ E2 , if they exist, are called partial derivatives of f at u0 ∈ U and are denoted by D1 f (u0 ) ∈ L(E1 , F), D2 f (u0 ) ∈ L(E2 , F). 2.4.12 Proposition.

Let U ⊂ E1 ⊕ E2 be open and f : U → F.

(i) If f is differentiable, then the partial derivatives exist and are given by D1 f (u) · e1 = Df (u) · (e1 , 0)

and

D2 f (u) · e2 = Df (u) · (0, e2 ).

(ii) If f is differentiable, then Df (u) · (e1 , e2 ) = D1 f (u) · e1 + D2 f (u) · e2 . (iii) f is of class C r iff Di f : U → L(Ei , F), i = 1, 2 both exist and are of class C r−1 . Proof. To prove (i), let ju1 : E1 → E1 ⊕ E2 be defined by ju1 (v1 ) = (v1 , u2 ), where u = (u1 , u2 ) . Then ju1 is C ∞ and Dju1 (u1 ) = J1 ∈ L(E1 , E1 ⊕ E2 ) is given by J1 (e1 ) = (e1 , 0). By the chain rule, D1 f (u) = D(f ◦ ju1 )(u1 ) = Df (u) ◦ J1 , which proves the first relation in (i). One similarly defines ju2 , J2 , and proves the second relation. Turning to (ii), let Pi (e1 , e2 ) = ei , i = 1, 2 be the canonical projections. Then compose the relation J1 ◦ P1 + J2 ◦ P2 = identity on E1 ⊕ E2 with Df (u) on the left and use (i). Finally we prove (iii). Let Φi ∈ L(L(E1 ⊕ E2 , F), L(Ei , F)) and Ψi ∈ L(L(Ei , F), L(E1 ⊕ E2 , F)) be defined by Φi (A) = A ◦ Ji and Ψi (Bi ) = Bi ◦ Pi , i = 1, 2. Then (i) and (ii) become Di f = Φi ◦ Df

Df = Ψ1 ◦ D1 f + Ψ2 ◦ D2 f.

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2. Banach Spaces and Differential Calculus

This shows that if f is differentiable, then f is C r iff D1 f and D2 f are C r−1 . Thus to conclude the proof we need to show that if D1 f and D2 f exist and are continuous, then Df exists. By Proposition 2.4.7 applied consecutively to the two arguments, we get f (u1 + e1 , u2 + e2 ) − f (u1 , u2 ) − D1 f (u1 , u2 ) · e1 − D2 f (u1 , u2 ) · e2 = f (u1 + e1 , u2 + e2 ) − f (u1 , u2 + e2 ) − D1 f (u1 , u2 ) · e1 + f (u1 , u2 + e2 ) − f (u1 , u2 ) − D2 f (u1 , u2 ) · e2  Z 1 (D1 f (u1 + te1 , u2 + e2 ) − D1 f (u1 , u2 )) dt · e1 = 0 1

Z +

 (D2 f (u1 , u2 + te2 ) − D2 f (u1 , u2 )) dt · e2 .

0

Taking norms and using in each term the obvious inequality ke1 k ≤ ke1 k + ke2 k ≡ k(e1 , e2 )k, we see that kf (u1 + e1 , u2 + e2 ) − f (u1 , u2 ) − D1 f (u1 , u2 ) · e1 − D2 f (u1 , u2 ) · e2 k  ≤ sup kD1 f (u1 + te1 , u2 + e2 ) − D1 f (u1 , u2 + e2 )k 0≤t≤1

 + sup kD2 f (u1 , u2 + te2 ) − D2 f (u1 , u2 )k k(e1 , e2 )k. 0≤t≤1

Both sups in the parentheses converge to zero as (e1 , e2 ) → (0, 0) by continuity of the partial derivatives.  Higher Derivatives.

If E1 = E2 = R and {e1 , e2 } is the standard basis in R2 we see that f (x + h, y) − f (x, y) ∂f (x, y) = lim = D1 f (x, y) · e1 ∈ F. h→0 ∂x h

Similarly, (∂f /∂y)(x, y) = D2 f (x, y) · e2 ∈ F. Define inductively higher derivatives ∂2f ∂ = ∂x2 ∂x 2.4.13 Example. R2 → R



∂f ∂x

 ,

∂ ∂2f = ∂x∂y ∂y



∂f ∂x

 ,

etc.

As an application of the formalism just introduced we shall prove that for f : U ⊂ 2 2 ∂2f 1 2 ∂ f 2 1 ∂ f (u) + v w (u) + v w (u) ∂x2 ∂y∂x ∂x∂y ∂2f + v 2 w2 2 (u), ∂y  2  ∂ f ∂2f  1 (u) (u)  ∂x2 ∂y ∂x   w2 , = (v 1 , v 2 )   ∂2f  w ∂2f (u) (u) ∂x ∂y ∂y 2

D2 f (u) · (v, w) = v 1 w1

where u ∈ U , v, w ∈ R2 , v = v 1 e1 + v 2 e2 , w = w1 e1 + w2 e2 , and {e1 , e2 } is the standard basis of R2 . To prove this, note that by definition, D2 f (u) · (v, w) = D((Df )(·) · w)(u) · v.

2.4 Properties of the Derivative

81

Applying the chain rule to Df (·) · w = Tw : A ∈ L(R2 , F) 7→ A · w ∈ F, the preceding expression becomes D(Df (·) · w)(u) · v = D(D1 f (·) · w1 e1 + D2 f (·) · w2 e2 )(u) · v (by Prop. 2.4.12(ii))   ∂f ∂f + w2 (u) · v = D w1 ∂x ∂y       ∂f ∂f = w1 D1 (u) · v 1 e1 + D2 (u) · v 2 e2 ∂x ∂x       ∂f ∂f 2 1 2 + w D1 (u) · v e1 + D2 (u) · v e2 ∂y ∂y ∂2f ∂2f ∂2f ∂2f (u) + v 1 w2 (u) + v 2 w2 2 (u). = v 1 w1 2 (u) + v 2 w1 ∂x ∂x∂y ∂y∂x ∂y

(2.4.1)



For computation of higher derivatives, note that by repeated application of Proposition 2.4.6, ( !) r X d d r D f (u) · (e1 , . . . , er ) = ··· f u+ ti ei . dtr dt1 i=1 t1 =···=tr =0

In particular, for f : U ⊂ Rm → Rn the components of Dr f (u) in terms of the standard basis are ∂xi1

∂rf , · · · ∂xir

0 ≤ ik ≤ r.

Thus, f is of class C r iff all its r-th order partial derivatives exist and are continuous. Symmetry of Higher Derivatives. Equality of mixed partials is of course a fundamental property we learn in calculus. Here is the general result. 2.4.14 Proposition (L. Euler). symmetric.

If f : U ⊂ E → F is C r , then Dr f (u) ∈ Lrs (E, F); that is, Dr f (u) is

Proof. First we prove the result for r = 2. Let u ∈ U , v, w ∈ E be fixed; we want to show that D2 f (u) · (v, w) = D2 f (u) · (w, v). To this, define the linear map a : R2 → E by a(e1 ) = v, and a(e2 ) = w, where e1 and e2 are the standard basis vectors of R2 . For (x, y) ∈ R2 , then a(x, y) = xv + yw. Now define the affine map A : R2 → E by A(x, y) = u + a(x, y). Since D2 (f ◦ A)(0, 0) · (e1 , e2 ) = D2 f (u) · (v, w) (Exercise 2.3-5), it suffices to prove the following formula: D2 (f ◦ A) · (x, y) · (e1 , e2 ) = D2 (f ◦ A)(x, y) · (e2 , e1 ); that is, ∂ 2 (f ◦ A) ∂ 2 (f ◦ A) = ∂x∂y ∂y∂x (see Example 2.4.13). Let g = f ◦ A : V = A−1 (U ) ⊂ R2 → F. Since for any λ ∈ F∗ , ∂ 2 (λ ◦ g)/∂x∂y = λ(∂ 2 g/∂x∂y), using the Hahn–Banach theorem 2.2.12, it suffices to prove that ∂2ϕ ∂2ϕ = , ∂x∂y ∂y∂x

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2. Banach Spaces and Differential Calculus

where ϕ = λ ◦ g : V ⊂ R2 → R, which is a standard result from calculus. For the sake of completeness we recall the proof. Applying the mean value theorem twice, we get Sh,k = [ϕ(x + h, y + k) − ϕ(x, y + k)] − [ϕ(x + h, y) − ϕ(x, y)]   ∂ϕ ∂ϕ (ch,k , y + k) − (ch,k , y) k = ∂x ∂x ∂2ϕ = (ch,k , dh,k )hk. ∂x∂y for some ch,k , dh,k lying between x and x + h, and y and y + k, respectively. By interchanging the two middle terms in Sh,k we can derive in the same way that Sh,k =

∂2ϕ (γh,k , δh,k )hk. ∂y∂x

Equating these two formulas for Sh,k , canceling h, k, and letting h → 0, k → 0, the continuity of D2 ϕ gives the result. For general r, proceed by induction: Dr f (u) · (v1 , v2 , . . . , vn ) = D2 (Dr−2 f )(u) · (v1 , v2 ) · (v3 , . . . , vn ) = D2 (Dr−2 f )(u) · (v2 , v1 ) · (v3 , . . . , vn ) = Dr f (u) · (v2 , v1 , v3 , . . . , vn ). Let σ be any permutation of {2, . . . , n}, so by the inductive hypothesis Dr−1 f (u)(v2 , . . . , vn ) = Dr−1 f (u)(vσ(2) , . . . , vσ(n) ). Take the derivative of this relation with respect to u ∈ U keeping v2 , . . . , vn fixed and get (Exercise 2.4-6): Dr f (u)(v1 , . . . , vn ) = Dr f (u)(v1 , vσ(2) , . . . , vσ(n) ). Since any permutation can be written as a product of the transposition {1, 2, 3, . . . , n} → {2, 1, 3, . . . , n} (if necessary) and a permutation of the set {2, . . . , n}, the result follows.  Taylor’s Theorem. Suppose U ⊂ E is an open convex set. Since + : E × E → E is continuous, there ˜ ⊂ E × E with these three properties: exists an open set U ˜, (i) U × {0} ⊂ U ˜ and 0 ≤ ξ ≤ 1, and (ii) u + ξh ∈ U for all (u, h) ∈ U ˜ implies u ∈ U . (iii) (u, h) ∈ U ˜ a thickening of U . For example, the choice Let us call such a set U ˜default = {(+)−1 (U )} ∩ (U × E) = {(u, h) ∈ U × E | u + h ∈ U } U is the default thickening. See Figure 2.4.2 for the default thickening and another choice of thickening of an open interval on R. ˜ be a thickening of 2.4.15 Theorem (Taylor’s Theorem). Let U be an open convex subset of E and let U r U . A map f : U ⊂ E → F is of class C iff there are continuous mappings ϕp : U ⊂ E → Lps (E, F),

p = 1, . . . , r,

and

˜ → Lrs (E, F), R:U

2.4 Properties of the Derivative

83

E

E

U

Figure 2.4.2. Two examples of thickeneings of an open interval on the real line.

˜, such that for all (u, h) ∈ U f (u + h) = f (u) +

ϕ1 (u) ϕ2 (u) 2 ϕr (u) r ·h+ · h + ··· + · h + R(u, h) · hr , 1! 2! r!

where hp = (h, . . . , h) (p times) and R(u, 0) = 0. If f is C r then necessarily ϕp = Dp f for all p = 1, . . . , r and, in addition, Z 1 (1 − t)r−1 r (D f (u + th) − Dr f (u)) dt. R(u, h) = (r − 1)! 0 Proof. We shall prove the “only if” part. The converse is proved in Supplement 2.4B. Leibniz’ rule gives the following integration by parts formula. If [a, b] ⊂ U ⊂ R and ψi : U ⊂ R → Ei , i = 1, 2 are C 1 mappings and B ∈ L(E1 , E2 ; F) is a bilinear map of E1 × E2 to F, then Z

b

B(ψ10 (t), ψ2 (t)) dt = B(ψ1 (b), ψ2 (b)) − B(ψ1 (a), ψ2 (a))

a

Z −

b

B(ψ1 (t), ψ20 (t)) dt.

a

Assume f is a C r mapping. If r = 1, then by Proposition 2.4.7 Z 1  f (u + h) = f (u) + Df (u + th) dt · h 0

Z = f (u) + Df (u) · h +

1

 (Df (u + th) − Df (u)) dt · h

0

and the formula is proved. For general k ≤ r proceed by induction choosing in the integration by parts formula E1 = R, E2 = E, B(s, e) = se, ψ2 (t) = Dk f (u + th) · hk , and ψ1 (t) = −(1 − t)k /k!, and taking into

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account that Z

1

0 k

Since D f (u) ∈

Lks (E, F)

(1 − t)k 1 dt = . k! (k + 1)!

by Proposition 2.4.14, Taylor’s formula follows.



Note that R(u, h) · hr = o(hr ) since R(u, h) → 0 as h → 0. If f is C r+1 then the mean value inequality and a bound on Dr+1 f gives R(u, h) · hr = o(hr+1 ). See Exercise 2.4-13 for the differentiability of R. The proof also shows that Taylor’s formula holds if f is (r − 1) times differentiable on U and r times differentiable at u. The estimate R(u, h) · hr = o(hr ) is proved directly by induction; for r = 1 it is the definition of the Fr´echet derivative. If f is C ∞ (i.e., is C r for all r) then we may be able to extend Taylor’s formula into a convergent power series. If we can, we say f is of class C ω , or analytic. A standard example of a C ∞ function that is not analytic is the following function from R to R (Figure 2.4.3) o n ( 1 , |x| < 1; exp − 1−x 2 θ(x) = 0, |x| ≥ 1.

y

1 x

–1 Figure 2.4.3. A bump function

This function is C ∞ , and all derivatives are 0 at x = ±1. To see this note that for |x| < 1,   −1 f (n) (x) = Qn (x)(1 − x2 )−2n exp , 1 − x2 where Qn (x) are polynomials given recursively by Q0 (x) = 1,

Qn+1 (x) = (1 − x2 )2 Q0n (x) + 2x(2n − 1 − 2nx2 )Qn (x).

Hence all coefficients of the Taylor series around these points vanish. Since the function is not identically 0 in any neighborhood of ±1, it cannot be analytic there. 2.4.16 Example (Differentiating Under the Integral). Let U ⊂ E be open and f : [a, b] × U → F. For t ∈ [a, b], define g(t) : U → F by g(t)(u) = f (t, u). If, for each t, g(t) is of class C r and if the maps (t, u) ∈ [a, b] × U 7→ Dj (g(t))(u) ∈ Ljs (E, F) are continuous, then h : U → F, defined by b

Z h(u) =

Z f (t, u) dt =

a

is C r and Dj h(u) =

Z a

b

g(t)(u) dt a

b

Dju f (t, u) dt,

j = 1, . . . , r,

2.4 Properties of the Derivative

85

where Du means the partial derivative in u. For r = 1, write

Z b



D(g(t))(u) · e dt

h(u + e) − h(u) −

a

Z Z 

b 1

(D(g(t))(u + se) · e − D(g(t))(u) · e) ds dt =

a 0 ≤ (b − a)kek

sup

kD(g(t))(u + se) − D(g(t))(u)k = o(e).

a≤t≤b,0≤s≤1

For r > 1 one can also use an argument like this, but the converse to Taylor’s theorem also yields the result rather easily. Indeed, if R(t, u, e) denotes the remainder for the C r Taylor expansion of g(t), then with ϕp = Dp h =

b

Z

Dp [g(t)] dt,

a

Rb the remainder for h is clearly R(u, e) = a R(t, u, e) dt. But R(t, u, e) → 0 as e → 0 uniformly in t, so R(u, e) is continuous and R(u, 0) = 0. Thus h is C r .  Extrema for Real Valued Functions on Banach Spaces. parallel to calculus.

Much of this theory proceeds in a manner

2.4.17 Definition. Let f : U ⊂ E → R be a continuous function, U open in E. We say f has a local minimum (resp., maximum) at u0 ∈ U , if there is a neighborhood V of u0 , V ⊂ U such that f (u0 ) ≤ f (u) (resp., f (u0 ) ≥ f (u)) for all u ∈ V . If the inequality is strict, u0 is called a strict local minimum (resp., maximum). The point u0 is called a global minimum (resp., global maximum) if f (u0 ) ≤ f (u) (resp., f (u0 ) ≥ f (u)) for all u ∈ U . Local maxima and minima are called local extrema. 2.4.18 Proposition. Let f : U ⊂ E → R be a continuous function differentiable at u0 ∈ U . If f has a local extremum at u0 , then Df (u0 ) = 0. Proof. If u0 is a local minimum, then there is a neighborhood V of U such that f (u0 + th) − f (u0 ) ≥ 0 for all h ∈ V . Therefore, the limit of [f (u0 + th) − f (u0 )]/t as t → 0, t ≥ 0 is ≥ 0 and as t → 0, t ≤ 0 is ≤ 0. Since both limits equal Df (u0 ), it must vanish.  This criterion is not sufficient as the elementary calculus example f : R → R, f (x) = x3 shows. Also, if U is not open, the values of f on the boundary of U must be examined separately. 2.4.19 Proposition.

Let f : U ⊂ E → R be twice differentiable at u0 ∈ U .

(i) If u0 is a local minimum (maximum), then D2 f (u0 ) · (e, e) ≥ 0 (≤ 0) for all e ∈ E. (ii) If u0 is a non-degenerate critical point of f , that is, Df (u0 ) = 0 and D2 f (u0 ) defines an isomorphism of E with E∗ , and if D2 f (u0 ) · (e, e) > 0 (< 0) for all e 6= 0, e ∈ E, then u0 is a strict local minimum (maximum) of f . Proof.

To prove (i), we note that by Taylor’s formula, in a neighborhood V of u0 , we have 0 ≤ f (u0 + h) − f (u0 ) =

1 2 D f (u0 )(h, h) + o(h2 ) 2

for all h ∈ V . If e ∈ E is arbitrary, for small t ∈ R, te ∈ V , so that 0≤

1 2 D f (u0 )(te, te) + o(t2 e2 ) 2

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2. Banach Spaces and Differential Calculus

implies D2 f (u0 )(e, e) +

2 o(t2 e2 ) ≥ 0. t2

Now let t → 0 to get the result. To prove (ii), denote by T : E → E∗ the isomorphism defined by e 7→ D2 f (u0 ) · (e, ·), so that there exists a > 0 such that akek ≤ kT ek = sup |hT e, e0 i| = sup |D2 f (u0 ) · (e, e0 )|. ke0 k=1

ke0 k=1

By hypothesis and symmetry of the second derivative, 0 < D2 f (u0 ) · (e + se0 , e + se0 ) = s2 D2 f (u0 ) · (e0 , e0 ) + 2sD2 f (u0 ) · (e, e0 ) + D2 f (u0 ) · (e, e), which is a quadratic form in s. Therefore, its discriminant must be negative, that is, |D2 f (u0 ) · (e, e0 )|2 < D2 f (u0 ) · (e0 , e0 )D2 f (u0 ) · (e, e) ≤ kD2 f (u0 )kD2 f (u0 ) · (e, e), and thus, we get akek ≤ sup |D2 f (u0 ) · (e, e0 )| ≤ kD2 f (u0 )k1/2 [D2 f (u0 ) · (e, e)]1/2 . ke0 k=1

Therefore, letting m = a2 /kD2 f (u0 )k, the following inequality holds for any e ∈ E: D2 f (u0 ) · (e, e) ≥ mkek2 . Thus, by Taylor’s theorem we have f (u0 + h) − f (u0 ) =

mkhk2 1 2 D f (u0 ) · (h, h) + o(h2 ) ≥ + o(h2 ). 2 2

Let ε > 0 be such that if khk < ε, then |o(h2 )| ≤ mkhk2 /4, which implies f (u0 + h) − f (u0 ) ≥ mkhk2 /4 > 0 for h 6= 0, and thus u0 is a strict local minimum of f .  The condition in (i) is not sufficient for f to have a local minimum at u0 . For example, f : R2 → R, f (x, y) = x2 − y 4 has f (0, 0) = 0, Df (0, 0) = 0, D2 f (0, 0) · (x, y)2 = 2x2 ≥ 0 and in any neighborhood of the origin, f changes sign. The conditions in (ii) are not necessary for f to have a strict local minimum at u0 . For example f : R → R, f (x) = x4 has f (0) = f 0 (0) = f 00 (0) = f 000 (0) = 0, f (4) (0) > 0 and 0 is a strict global minimum for f . On the other hand, if the conditions in (ii) hold and u0 is the only critical point of a differentiable function f : U → R, one might think that u0 is a strict global minimum of f . While true in one dimension by Rolle’s theorem, this is not true in general, as the C ∞ function f : R2 → R defined by p 2 f (x, y) = −y 4 − e−x + 2y 2 ex + e−x2 shows. Indeed, (0, 0) is readily seen to be the only critical point and at this point, all the hypotheses of √ Proposition 2.4.19 are satisfied (the matrix of D2 f (0, 0) has the only non zero entries 2 and 4 2 on the diagonal). Thus, the function √ f has a unique strict local minimum at the origin whose value is f (0, 0) = −1. However, f (0, 2) = −17 + 8 2 < −1 and hence the origin is not a global minimum of f . Also, care has to be taken with the statement in (ii): non-degeneracy holds in the topology of E. If E is continuously embedded in another Banach space F and D2 f (u0 ) is non-degenerate in F only, u0 need not even be a minimum. For example, consider the smooth map Z 1 1 f : L4 ([0, 1]) → R, f (u) = (u(x)2 − u(x)4 ) dx 2 0

2.4 Properties of the Derivative

87

and note f (0) = 0, Df (0) = 0, and Z

2

1

v(x)2 dx > 0

D f (0)(v, v) =

for v 6= 0,

0

and that D2 f (0) defines an isomorphism of L4 ([0, 1]) with L4/3 ([0, 1]). Alternatively, D2 f (0) is non-degenerate on L2 ([0, 1]) not on L4 ([0, 1]). Also note that in any neighborhood of 0 in L4 ([0, 1]), f changes sign: f (1/n) = (n2 − 1)/2n4 ≥ 0 for n ≥ 2, but f (un ) = −6/n < 0 for n ≥ 1 if ( 2, on [0, 1/n]; un = 0, elsewhere and both 1/n, un converge to 0 in L4 ([0, 1]). Thus, even though D2 f (0) is positive, 0 is not a minimum of f . (See Ball and Marsden [1984] for more sophisticated examples of this sort.)

Supplement 2.4A The Leibniz and Chain Rules Here the explicit formulas are given for the kth order derivatives of products and compositions. The proofs are straightforward but quite messy induction arguments, which will be left to the interested reader. The Higher Order Leibniz Rule. Let E, F1 , F2 , and G be Banach spaces, U ⊂ E an open set, f : U → F1 and g : U → F2 of class C k and B ∈ L(F1 , F2 ; G). Let f × g : U → F1 × F2 denote the mapping (f × g)(e) = (f (e), g(e)) and let B(f, g) = B ◦ (f × g). Thus B(f, g) is of class C k and by Leibniz’ rule, DB(f, g)(p) · e = B(Df (p) · e, g(p)) + B(f (p), Dg(p) · e). Higher derivatives of f and g are maps Di f : U → Li (E; F1 ),

Dk−i g : U → Lk−i (E; F2 ),

where D0 f = f,

D0 g = g,

L0 (E; F1 ) = F1 ,

L0 (E; F2 ) = F2 .

Denote by λi,k−i ∈ L(Li (E; F1 ), Lk−i (E, F2 ); Lk (E; G)), the bilinear mapping defined by [λi,k−i (A1 , A2 )](e1 , . . . , ek ) = B(A1 (e1 , . . . , ei ), A2 (ei+1 , . . . , ek )) for A1 ∈ Li (E; F1 ), A2 ∈ Lk−i (E; F2 ), and e1 , . . . , ek ∈ E. Then λi,k−i (Di f, Dk−i g) : U → Lk (E; G) is defined by λi,k−i (Di f, Dk−i g)(p) = λi,k−i (Di f (p), Dk−i g(p)) for p ∈ U . Leibniz’ rule for kth derivatives is Dk B(f, g) = Symk ◦

k   X k i=0

i

λi,k−i (Di f, Dk−i g),

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2. Banach Spaces and Differential Calculus

where Symk : Lk (E; G) → Lks (E; G) is the symmetrization operator, given by (see Exercise 2.2-9): 1 X (Symk A)(e1 , . . . , ek ) = A(eσ(1) , . . . , e(k) ), k! σ∈Sk

where Sk is the group of permutations of {1, . . . , k}. Explicitly, taking advantage of the symmetry of higher order derivatives, this formula is Dk B(f, g)(p) · (e1 , . . . , ek ) k   XX k = B(Di f (p) · (eσ(1) , . . . , eσ(i) ), Dk−i g(p)(eσ(i+1) , . . . , eσ(k) )), i σ

i=0

where the outer sum is over all permutations σ ∈ Sk such that σ(1) < · · · < σ(i)

and σ(i + 1) < · · · < σ(k).

The Higher Order Chain Rule. Let E, F, and G be Banach spaces and U ⊂ F and V ⊂ F be open sets. Let f : U → V and g : V → G be maps of class C k . By the usual chain rule, g ◦ f : U → G is of class C k and D(g ◦ f )(p) = Dg(f (p)) ◦ Df (p) for p ∈ U . For every tuple (i, j1 , . . . , ji ), where i > 1, and j1 + · · · + ji = k, define the continuous multilinear map λi,j1 ,...,ji : Li (F; G) × Lj1 (E; F) × · · · × Lji (E; F) → Lk (E; G) by λi,j1 ,...,ji (A, B1 , . . . , Bi ) · (e1 , . . . , ek ) = A(B1 (e1 , . . . , ej1 ), B2 (ej1 +1 , . . . , ej1 +j2 ), . . . , Bi (eji +···+ji−1 +1 , . . . , ek )) for A ∈ Li (F; G),

B` ∈ Lj` (E; F),

` = 1, . . . , i and ei , . . . , ek ∈ E.

Since Dj` f : U → Lj` (E; F), we can define λi,j1 ,...,ji ◦ (Di g ◦ f × Dj1 f × · · · × Dji ) : U → Lk (E; G) by p 7→ λi,j1 ,...,ji (Di g(f (p)), Dj1 f (p), . . . , Dji f (p)). With these notations, the kth order chain rule is Dk (g ◦ f ) = Symk ◦

k X

X

i=1 j1 +···+ji =k

k! λi,j1 ,...,ji j1 ! · · · ji !

◦ (Di g ◦ f × Dj1 f × · · · × Dji f ), where Symk : Lk (E; G) → Lks (E; G) is the symmetrization operator. Taking into account the symmetry of higher order derivatives, the explicit formula at p ∈ U and e1 , . . . , ek ∈ E, is Dk (g ◦ f )(p) · (e1 , . . . , ek ) =

k X

X

X

Di g(f (p))(Dj1 f (p) · (e`1 , . . . , e`j1 ), . . . ,

i=1 j1 +···+ji =k

Dji f (p) · (e`j1 +···+ji−1 +1 , . . . , e`k )) where the third sum is taken over indices satisfying `1 < · · · < `j1 , · · · , `j1 +···+ji−1 +1 < · · · < `k .

2.4 Properties of the Derivative

89

Supplement 2.4B The Converse to Taylor’s Theorem This theorem goes back to Marcinkiewicz and Zygmund [1936], Whitney [1943a], and Glaeser [1958]. The proof of the converse that we shall follow is due to Nelson [1969]. Assume the formula in the theorem holds where ϕp = Dp f , 1 ≤ p ≤ r, and that R(u, h) has the desired expression. If r = 1, the formula reduces to the definition of the derivative. Hence ϕ1 = Df , f is C 1 , and thus R(u, h) has the desired form, using Proposition 2.4.7. Inductively assume the theorem is true for r = p − 1. Thus ϕj = Dj f , for 1 ≤ j ≤ p − 1. Let h, k ∈ E be small in norm such that u + h + k ∈ U . Write the formula in the theorem for f (u + h + k) in two different ways: f (u + h + k) = f (u + h) + Df (u + h) · k + · · · 1 Dp−1 f (u + h) · k p−1 + (p − 1)! 1 + ϕp (u + h) · k p + R1 (u + k, k) · k p ; p! f (u + h + k) = f (u) + Df (u) · (h + k) + · · · 1 + Dp−1 f (u) · (h + k)p−1 (p − 1)! 1 + ϕp (u) · (h + k)p + R2 (u, h + k) · (h + k)p . p! Subtracting them and collecting terms homogeneous in k j we get: g0 (h) + g1 (h) · k + · · · + gp−1 (h) · k p−1 + gp (h) · k p = R1 (u + h, k) · k p − R2 (u, h + k) · (h + k)p , where gj (h) ∈ Lj (E; F) is given by " p−1−j X 1 1 Dj f (u + h) − Dj f (u) − Dj+i f (u) · hi gj (h) = j! i! i=1 # 1 − ϕp (u) · hp−j , (p − j)! for 0 ≤ j ≤ p − 2; gp−1 (h) =

 p−1  1 D f (u + h) − Dp−1 f (u) − ϕp (u) · h ; (p − 1)!

and gp (h) =

1 [ϕp (u + h) − ϕp (u)] . p!

Note that gj (0) = 0 and that the maps gj are continuous. Let kkk satisfy (1/4)khk ≤ kkk ≤ (1/2)khk. Since kR1 (u + h, k) · k p − R2 (u, h + k) · (h + k)p − gp (h) · k p k ≤ (kR1 (u + h, k)k + kgp (h)k) kkkp + kR2 (u, h + k)k(khk + kkk)p ≤ {kR1 (u + h, k)k + kgp (h)k + kR2 (u, h + k)k} (1 + 3p )khkp /2p

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2. Banach Spaces and Differential Calculus

and the quantity in braces {} → 0 as h → 0, it follows that R1 (u + h, k) · k p − R2 (u, h + k) · (h + k)p − gp (h) · k p = o(hp ). Hence g0 (h) + g1 (h) · k + · · · + gp−1 (h) · k p−1 = o(hp ). We claim that subject to the condition (1/4)khk ≤ kkk ≤ (1/2)khk, each term of this sum is o(hp ). If λ1 , . . . , λp are distinct numbers, replace k by λj k in the foregoing, and get a p × p linear system in the unknowns g0 (h), . . . , gp−1 (h) · k p−1 with Vandermonde determinant Πi<j (λi − λj ) 6= 0 and right-hand side a column vector all of whose entries are o(hp ). Solving this system we get the result claimed. In particular, (Dp−1 f (u + h) − Dp−1 f (u) − ϕp (u) · h) · k p−1 = gp−1 (h) · k p−1 = o(hp ). Using polarization (see Supplement 2.2B) we get kDp−1 f (u + h) − Dp−1 f (u) − ϕp (u) · hk

(p − 1)p−1 p−1 p−1 0

(D f (u + h) − D f (u) − ϕp (u) · h) ≤ (p − 1)!

p−1 (p − 1)p−1 (D f (u + h) − Dp−1 f (u) = sup (p − 1)! kek≤1

− ϕp (u) · h) · ep−1

p−1 (p − 1)p−1 = (D f (u + h) − Dp−1 f (u) sup (p − 1)! kkk≤khk/2  p−1

2k

− ϕp (u) · h) ·

khk

p−1 (2(p − 1))p−1

(D f (u + h) − Dp−1 f (u) = sup (p − 1)!khkp−1 kkk≤khk/2

− ϕp (u) · h) · k p−1

=

(2(p − 1))p−1 o(hp ). (p − 1)!khkp−1

Since o(hp )/khkp → 0 as h → 0, this relation proves that Dp−1 f is differentiable and Dp f (u) = ϕp (u). Thus f is of class C p , ϕp being continuous, and the formula for R follows by subtracting the given formula for f (u + h) from Taylor’s expansion.  The converse of Taylor’s theorem provides an alternative proof that Dr f (u) ∈ Lrs (E; F). Observe first that in the proof of Taylor’s expansion for a C r map f the symmetry of Dj f (u) was never used, so if one symmetrizes the Dj f (u) and calls them ϕj , the same expansion holds. But then the converse of Proposition 2.4.12 says that ϕj = Dj f . We shall consider here simple versions of two theorems from global analysis, which shall be used in Supplement 4.1C, namely the smoothness of the evaluation mapping and the “omega lemma.” The Evaluation Map. Let I = [0, 1] and E be a Banach space. The vector space C r (I; E) of C r -maps (r > 0) of I into E is a Banach space with respect to the norm kf kr = max sup kDi f (t)k 1≤i≤r t∈I

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91

(see Exercise 2.4-8). If U is open in E, then the set C r (I; U ) = { f ∈ C r (I; E) | f (I) ⊂ U } is checked to be open in C r (I; E). 2.4.20 Proposition.

The evaluation map defined by: ev : C r (I; U ) × ]0, 1[ → U

defined by ev(f, t) = f (t) is C r and its kth derivative, k = 0, 1, . . . , r, is given by Dk ev(f, t) · ((g1 , s1 ), . . . , (gk , sk )) = Dk f (t) · (s1 , . . . , sk ) +

k X

Dk−1 gi (t) · (s1 , . . . , si−1 , si+1 , . . . , sk )

i=1

where (gi , si ) ∈ C r (I; E) × R,

i = 1, . . . , k.

Proof. For (g, s) ∈ C r (I; E) × R, define the norm k(g, s)k = max(kgkr , |s|). Note that the right-hand side of the formula in the statement is symmetric in the arguments (gi , si ), i = 1, . . . , k. We shall let this right-hand side be denoted ϕk : C r (I; U ) × ]0, 1[ → Lks (C r (I; E) × R; E),

k = 1, . . . , r,

and we set ϕ0 (f, t) = f (t). The proposition holds for r = 0 by uniform continuity of f on I since kf (t) − g(s)k ≤ kf (t) − f (s)k + kf − gk0 . Since

Dr g(t) · sr =0 (g,s)→(0,0) k(g, s)kr lim

for all t ∈ ]0, 1[, by Taylor’s theorem for f and g we get ev(f + g, t + s) = f (t + s) + g(t + s) r X 1 i = (D f (t) · si + Di g(t) · si ) + R(t, s) · sr i! i=0 =

r X 1 ϕi (f, t) · (g, s)i + R((f, t), (g, s)) · (g, s)r , i! i=0

where R((f, t), (g, s)) · ((g1 , s1 ), . . . , (gr , sr )) = R(t, s) · (s1 , . . . , sr ) r 1 X r D gi (t) · (s1 , . . . , sr ), + rr! i=1 which is symmetric in its arguments and R((f, t), (0, 0)) = 0. By the converse to Taylor’s theorem, the proposition is proved if we show that every ϕi , 1 ≤ i ≤ r, is continuous. Since kDk−1 gi (t) − Dk−1 gi (s)k ≤ |t − s| sup kDk gi (u)k ≤ |t − s| kgi kr u∈I

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by the mean value theorem, the inequality k(ϕk (f, t) − ϕk (g, s)) · ((g1 , s1 ), . . . , (gk , sk ))k ≤ kDk f (t) − Dk g(s)k |s1 | · · · |sk | +

k X

kDk−1 gi (t) − Dk−1 gi (s)k |s1 | · · · |si−1 | |si+1 | · · · |sk |

i=1

implies kϕk (f, t) − ϕk (g, s)k ≤ kDk f (t) − Dk g(s)k + k|t − s| ≤ kDk f (t) − Dk f (s)k + kDk f (s) − Dk g(s)k + k|t − s| ≤ kDk f (t) − Dk f (s)k + 2kk(f, t) − (g, s)k. Thus the uniform continuity of Dk f on I implies the continuity of ϕk at (f, t).



Omega Lemma. We use the expression “omega Lemma” following terminology of Abraham [1963]. Various results of this type can be traced back to earlier works of Sobolev [1939] and Eells [1958]. Let M be a compact topological space and E, F be Banach spaces. With respect to the norm kf k = sup kf (m)k, m∈M

the vector space C 0 (M, E) of continuous E-valued maps on M , is a Banach space. If U is open in E, it is easy to see that C 0 (M, U ) = { f ∈ C 0 (M, E) | f (M ) ⊂ U } is open. 2.4.21 Proposition (Omega Lemma).

Let g : U → F be a C r map, r > 0. The map

Ωg : C 0 (M, U ) → C 0 (M, F)

defined by

Ωg (f ) = g ◦ f

is also of class C r . The derivative of Ωg is DΩg (f ) · h = [(Dg) ◦ f ] · h that is, [DΩg (f ) · h](x) = Dg(f (x)) · h(x). The formula for DΩg is quite plausible. Indeed, we have d d Ωg (f + εh)(x) = g(f (x) + εh(x)) . [DΩg (f ) · h](x) = dε dε ε=0 ε=0 By the chain rule this is Dg(f (x)) · h(x). This shows that if Ωg is differentiable, then DΩg must be as stated in the proposition. Proof.

Let f ∈ C 0 (M, U ). By continuity of g and compactness of M , kΩg (f ) − Ωg (f 0 )k = sup kg(f (m)) − g(f 0 (m))k m∈M

is small as soon as kf − f 0 k is small; that is, Ωg is continuous at each point f . Let Ai : C 0 (M, Lis (E; F)) → Lis (C 0 (M, E); C 0 (M, F))

2.4 Properties of the Derivative

93

be given by Ai (H)(h1 , . . . , hi )(m) = H(m)(h1 (m), . . . , hi (m)) 0

i

for H ∈ C (M, L (E; F)), h1 , . . . , hi ∈ C 0 (M, E) and m ∈ M . The maps Ai are clearly linear and are continuous with kAi k ≤ 1. Since Di g : U → Lis (E; F) is continuous, the preceding argument shows that the maps ΩDi g : C 0 (M, U ) → Lis (C 0 (M, E); C 0 (M, F)) are continuous and hence Ai ◦ ΩDi g : C 0 (M, U ) → Lis (C 0 (M, E); C 0 (M, F)) is continuous. The Taylor theorem applied to g yields g(f (m) + h(m)) = g(f (m)) +

r X 1 i D g(f (m)) · h(m)i i! i=1

+ R(f (m), h(m)) · h(m)i so that defining [(Di g ◦ f ) · hi ](m) = Di g(f (m)) · h(m)i , and [R(f, h) · (h1 , . . . , hr )](m) = R(f (m), h(m)) · (h1 (m), . . . , hr (m)) we see that R is continuous, R(f, 0) = 0, and Ωg (f + h) = g ◦ (f + h) = g ◦ f + = Ωg (f ) +

r X 1 i (D g ◦ f ) · hi + R(f, h) · hi i! i=1

r X 1 (Ai ◦ ΩDi g )(f ) · hi + R(f, h) · hi . i! i=1

Thus, by the converse of Taylor’s theorem, Di Ωg = Ai ◦ ΩDi g and Ωg is of class C r .



This proposition can be generalized to the Banach space C r (I, E), I = [a, b], equipped with the norm k · kr given by the maximum of the norms of the first r derivatives; that is, kf kr = max sup kf (i) (t)k. 0≤i≤r t∈I

If g is C r+q , then Ωg : C r (I, E) → C r−k (I, F) is C q+k . Readers are invited to convince themselves that the foregoing proof works with only trivial modifications in this case. This version of the omega lemma will be used in Supplement 4.1C. For applications to partial differential equations, the most important generalizations of the two previous propositions is to the case of Sobolev maps of class H s ; see for example Palais [1968], Ebin and Marsden [1970], and Marsden and Hughes [1983] for proofs and applications.

Supplement 2.4C The Functional Derivative and the Calculus of Variations Differential calculus in infinite dimensions has many applications, one of which is to the calculus of variations. We give some of the elementary aspects here to introduce the reader to the subject. We shall begin with some notation and a generalization of the notion of the dual space.

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Duality and Pairings. Let E and F be Banach spaces. A continuous bilinear functional h , i : E × F → R is called E-non-degenerate or E-weakly non-degenerate if hx, yi = 0 for all y ∈ F implies x = 0. Similarly, it is F-non-degenerate or F-weakly non-degenerate if hx, yi = 0 for all x ∈ E implies y = 0. If it is both, we just say h , i is (weakly) non-degenerate. Equivalently, the two linear maps of E to F∗ and F to E∗ defined by x 7→ hx, ·i and y 7→ h·, yi, respectively, are one to one. If they are isomorphisms, h , i is called E- or F-strongly non-degenerate. A non-degenerate bilinear form h , i thus represents certain linear functionals on F in terms of elements in E. We say E and F are in duality if there is a non-degenerate bilinear functional h , i : E × F → R, also called a pairing of E with F. If the functional is strongly non-degenerate, we say the duality is strong . 2.4.22 Examples. A. Let E = F∗ . Let h , i : F∗ × F → R be given by hϕ, yi = ϕ(y) so the map E → F∗ is the identity. Thus, h , i is E-strongly non-degenerate. It is easily checked that h , i is F-non-degenerate by making use of the Hahn–Banach theorem. (If it is F∗ strongly non-degenerate, F is called reflexive.) B. Let E = F and h , i : E × E → R be an inner product on E. Then h , i is non-degenerate since h , i is positive definite. If E is a Hilbert space, then h , i is a strongly non-degenerate pairing by the Riesz representation theorem.  Functional Derivatives. one defines the gradient.

We now define the functional derivative which uses the pairing similar to how

2.4.23 Definition. Let E and F be normed spaces and h , i be an E-weakly non-degenerate pairing. Let f : F → R be differentiable at the point α ∈ F. The functional derivative δf /δα of f with respect to α is the unique element in E, if it exists, such that   δf ,β for all β ∈ F. (2.4.2) Df (α) · β = δα Likewise, if g : E → R and h , i is F-weakly degenerate, we define the functional derivative δg/δv ∈ F, if it exists, by   δf Dg(v) · v 0 = v 0 , for all v 0 ∈ E. δv Weak non-degeneracy ensures the uniqueness of the functional derivatives, if they exist. In some interesting examples, E and F are spaces of mappings, as in the following example. 2.4.24 Example. Let Ω ∈ Rn be an open bounded set and consider the space E = C 0 (D), of continuous real valued functions on D where D = cl(Ω). Take F = C 0 (D) = E. The L2 -pairing on E × F is the bilinear map given by Z h , i : C 0 (D) × C 0 (D) → R,

f (x)g(x) dnx.

hf, gi = Ω

Let r be a positive integer and define f : E → R by Z 1 f (ϕ) = [ϕ(x)]r dnx. 2 Ω Then using the calculus rules from this section, we find Z Df (ϕ) · ψ = r[ϕ(x)]r−1 ψ(x) dnx. Ω

Thus,

δf = rϕr−1 . δϕ



2.4 Properties of the Derivative

95

Suppose, more generally, that f is defined on a Banach space E of functions ϕ on a region Ω in Rn . The functional derivative (δf /δϕ) of f with respect to ϕ is the unique element (δf /δϕ) ∈ E, if it exists, such that   Z   δf δf Df (ϕ) · ψ = ,ψ = (x)ψ(x) dnx for all ψ ∈ E. δϕ δϕ Ω The functional derivative may be determined in examples by Z d δf f (ϕ + εψ). (x)ψ(x) dnx = dε ε=0 Ω δϕ Criterion for Extrema.

(2.4.3)

A basic result in the calculus of variations is the following.

2.4.25 Proposition. Let E be a space of functions, as above. A necessary condition for a differentiable function f : E → R to have an extremum at ϕ is that δf = 0. δϕ Proof. If f has an extremum at ϕ, then for each ψ, the function h(ε) = f (ϕ + εψ) has an extremum at ε = 0. Thus, by elementary calculus, h0 (0) = 0. Since ψ is arbitrary, the result follows from equation (2.4.3).  Sufficient conditions for extrema in the calculus of variations are more delicate. See, for example, Bolza [1904] and Morrey [1966]. 2.4.26 Examples. A.

Suppose that Ω ⊂ R is an interval and that f , as a functional of ϕ ∈ C k (Ω), k ≥ 1, is of the form   Z dϕ dx (2.4.4) f (ϕ) = F x, ϕ(x), dx Ω

for some smooth function F : Ω × R × R → R, so that the right hand side of equation (2.4.4) is defined. We call F the density associated with f . It can be shown by using the results of the preceding supplement that f is smooth. Using the chain rule,   Z Z δf d d(ϕ + εψ) (x)ψ(x) dx = F x, ϕ + εψ, dx dε ε=0 Ω dx Ω δϕ   Z dϕ = D2 F x, ϕ(x), ψ(x) dx dx Ω   Z dϕ dψ dx, + D3 F x, ϕ(x), dx dx Ω where D2 F =

∂F ∂ϕ

and D3 F =

∂F ∂(∂ϕ/∂x)

denote the partial derivatives of F with respect to its second and third arguments. Integrating by parts, this becomes     Z  Z dϕ d dϕ D2 F x, ϕ(x), ψ(x) dx − D3 F x, ϕ(x), ψ(x) dx dx dx dx Ω Ω   Z dϕ + D3 F x, ϕ(x), ψ(x) dx. dx ∂Ω

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Let us now restrict our attention to the space of ψ’s which vanish on the boundary ∂Ω of Ω . In that case we get δf d = D2 F − D3 F. δϕ dx Rewriting this according to the designation of the second and third arguments of F as ϕ and dϕ/dx, respectively, we obtain δf ∂F d ∂F = − . (2.4.5) δϕ ∂ϕ dx ∂(dϕ/dx) By a similar argument, if Ω ⊂ Rn , equation (2.4.5) generalizes to δf ∂F ∂F d = − k . δϕ ∂ϕ dx ∂(dϕ/dxk )

(2.4.6)

(Here, a sum on repeated indices is assumed.) Thus, f has an extremum at ϕ only if d ∂F ∂F − k = 0. ∂ϕ dx ∂(∂ϕ/∂xk ) This is called the Euler–Lagrange equation in the calculus of variations. B. Assume that in Example A, the density F associated with f depends also on higher derivatives, that is, F = F (x, ϕ(x), ϕx , ϕxx , . . . ), where ϕx = dϕ/dx, ϕxx = d2ϕ/dx2 , etc. Therefore Z f (ϕ) = F (x, ϕ(x), ϕx , ϕxx , . . . ) dx. Ω

By an analogous argument, formula (2.4.5) generalizes to     ∂F d ∂F d2 ∂F δf = − + 2 − ··· δϕ ∂ϕ dx ∂ϕx dx ∂ϕxx

(2.4.7)

C. Consider a closed curve γ in R3 such that γ lies above the boundary ∂Ω of a region Ω in the xy-plane, as in Figure 2.4.4.

z

y

x Figure 2.4.4. A curve γ lying over ∂Ω

2.4 Properties of the Derivative

97

Consider differentiable surfaces in R3 (i.e., two-dimensional manifolds of R3 ) that are graphs of C k functions ϕ : Ω ⊂ R2 → R, so that (x, y, ϕ(x, y)) are coordinates on the surface. What is the surface of least area whose boundary is γ? From elementary calculus we know that the area as a function of ϕ is given by Z q 1 + ϕ2x + ϕ2y dx dy. A(ϕ) = Ω

From equation (2.4.6), a necessary condition for ϕ to minimize A is that ϕxx (1 + ϕ2y ) − 2ϕx ϕy ϕxy + ϕyy (1 + ϕ2x ) δA = 0, =− δϕ (1 + ϕ2x + ϕ2y )3/2

(2.4.8)

for (x, y) ∈ Ω . We relate this to the classical theory of surfaces as follows. From the differential geometry of surfaces, one learns that a surface has two principal curvatures κ1 and κ2 ; the mean curvature κ is defined to be their average: that is, κ = (κ1 + κ2 )/2. An elementary theorem of geometry asserts that κ is given by the formula ϕxx (1 + ϕ2y ) − 2ϕx ϕy ϕxy + ϕyy (1 + ϕ2x ) . (2.4.9) κ= (1 + ϕ2x + ϕ2y )1/2 If the surface represents a sheet of rubber, the mean curvature represents the net force due to internal stretching. Comparing equations (2.4.8) and (2.4.9) we find the that a minimal surface, that is, a surface with minimal area, has zero mean curvature.  Total Functional Derivative. Now consider the case in which f is a differentiable function of n variables, that is f is defined on a product of n function spaces Fi , i = 1, . . . , n; f : F1 × · · · × Fn → R and we have pairings h , ii : Ei × Fi → R. 2.4.27 Definition. The i-th partial functional derivative δf /δϕi of f with respect to ϕi ∈ Fi is defined by   δf d , ψi = f (ϕ1 , . . . , ϕi + εψi , . . . , ϕn ) δϕi dε ε=0 i = Di f (ϕ1 , . . . , ϕn ) · ψi = Df (ϕ1 , . . . , ϕn )(0, . . . , ψi , . . . , 0). (2.4.10) The total functional derivative is given by   δf , (ψ1 , . . . , ψn ) = Df (ϕ1 , . . . , ϕn ) · (ψ1 , . . . , ψn ) δ(ϕ1 , . . . , ϕn ) n X Di f (ϕ1 , . . . , ϕn )(0, . . . , ψi , . . . , 0) = i=1

 n  X δf = , ψi . δϕi i i=1 2.4.28 Examples. A. Suppose that f is a function of n functions ϕi ∈ C k (Ω), where Ω ⊂ Rn , and their first partial derivatives, and is of the form   Z ∂ϕi f (ϕ1 , . . . , ϕn ) = F x, ϕi , i dnx. ∂x Ω It follows that

δf ∂F ∂ ∂F  = − k  ∂ϕi δϕi ∂ϕi ∂x ∂ ∂xk

(sum on k)

(2.4.11)

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2. Banach Spaces and Differential Calculus

B. Classical Field Theory. As discussed in Goldstein [1980], Section 12 and in Marsden and Ratiu [1999], Chapter 3, Lagrange’s equations for a field η = η(x, t) with components η a follow from Hamilton’s variational principle. When the Lagrangian L is given by a Lagrangian density £, that is, L is of the form   ZZ a ∂η a j a ∂η L(η) = £ x ,η , j , dnx dt (2.4.12) ∂x ∂t Ω⊂R3 the variational principle states that η should be a critical point of L. Assuming appropriate boundary conditions, this results in the equations of motion 0=

d ∂ ∂£ ∂£ ∂£ δL = − a+ k a a δη dt ∂(∂η /∂t) ∂η ∂x ∂(∂η a /∂xk )

(2.4.13)

(a sum on k is understood). Regarding L as a function of η a and η˙a = ∂η a /∂t (rather than as a function of the pointwise values), the equations of motion take the form: d δL δL = a a ˙ dt δ η δη

(2.4.14)

C. Let Ω ⊂ Rn and let C∂k (Ω) stand for the C k functions vanishing on ∂Ω. Let f : C∂k (Ω) → R be given by the Dirichlet integral Z 1 f (ϕ) = h∇ϕ, ∇ϕi dnx, 2 Ω where h , i is the standard inner product on Rn . Differentiating with respect to ϕ: Z d 1 Df (ϕ) · ψ = h∇(ϕ + εψ), ∇(ϕ + εψ)i dnx dε ε=0 2 Ω Z = h∇ϕ, ∇ψi dnx Ω Z =− ∇2 ϕ(x) · ψ(x) dnx (integrating by parts). Ω

Thus δf /δϕ = −∇2 ϕ, the Laplacian of ϕ. D. The Stretched String. Consider a string of length ` and mass density σ, stretched horizontally under a tension τ , with ends fastened at x = 0 and x = `. Let u(x, t) denote the vertical displacement of the string at x, at time t. We have u(0, t) = u(`, t) = 0. The potential energy V due to small vertical displacements is shown in elementary mechanics texts to be  2 Z ` 1 ∂u V = τ dx, ∂x 0 2 and the kinetic energy T is Z T = 0

`

1 σ 2



∂u ∂t

2 dx.

From the definitions, we get ∂2u δV δT = −τ 2 and = σ u. ˙ δu ∂x δ u˙ Then with the Lagrangian L = T − V , the equations of motion (2.4.14) become the wave equation σ

∂2u ∂2u − τ 2 = 0. 2 ∂t ∂x



2.4 Properties of the Derivative

99

Next we formulate a chain rule for functional derivatives. Let h , i : E × F → R be a weakly nondegenerate pairing between E and F. If A ∈ L(F, F), its adjoint A∗ ∈ L(E, E), if it exists, is defined by hA∗ v, αi = hv, Aαi for all v ∈ E and α ∈ F. Let ϕ : F → F be a differentiable map and f : F → R be differentiable at α ∈ F. From the chain rule, D(f ◦ ϕ)(α) · β = Df (ϕ(α)) · (Dϕ(α) · β),

for β ∈ F.

Hence assuming that all functional derivatives and adjoints exist, the preceding relation implies       δf δ(f ◦ ϕ) ∗ δf ,β = , Dϕ(α) · β = Dϕ(α) · ,β δα δγ δγ where γ = ϕ(α), that is, δ(f ◦ ϕ) δf = Dϕ(α)∗ · . δα δγ

(2.4.15)

Similarly if ψ : R → R is differentiable then for α, β ∈ F, D(ψ ◦ f )(α) · β = Dψ(f (α)) · (Df (α) · β) where the first dot on the right hand side is ordinary multiplication by Dψ(f (a)) ∈ R. Hence       δf δf δ(ψ ◦ f ) 0 , β = Dψ(f (α)) · , β = ψ (f (α)) , β δα δα δα that is, δf δ(ψ ◦ f ) = ψ 0 (f (α)) . δα δα

(2.4.16)

Exercises  2.4-1. Show that if g : U ⊂ E → L(F, G) is C r , then f : U × F → G, defined by f (u, v) = (g(u))(v), u ∈ U , v ∈ F is also C r . Hint: Apply the Leibniz rule with L(F, G) × F → G the evaluation map.  2.4-2. Show that if f : U ⊂ E → L(F, G), g : U ⊂ E → L(G, H) are C r mappings then so is h : U ⊂ E → L(F, H), defined by h(u) = g(u) ◦ f (u).  2.4-3.

Extend Leibniz’ rule to multilinear mappings and find a formula for the derivative.

 2.4-4. Define a map f : U ⊂ E → F to be of class T 1 if it is differentiable, its tangent map T f : U × E → F × F is continuous and kDf (x)k is locally bounded. (i) For E and F finite dimensional, show that this is equivalent to C 1 . (ii) (Project.) Investigate the validity of the chain rule and Taylor’s theorem for T r maps . (iii) (Project.) Show that the function developed in Smale [1964] is T 2 but is not C 2 .  2.4-5. Suppose that f : E → F (where E, F are real Banach spaces) is homogeneous of degree k (where k is a nonnegative integer); that is, f (te) = tk f (e) for all t ∈ R, and e ∈ E. (i) Show that if f is differentiable, then Df (u) · u = kf (u). Hint: Let g(t) = f (tu) and compute dg/dt.

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(ii) If E = Rn and F = R, show that this relation is equivalent to n X

xi

i=1

∂f = kf (x) ∂xi

Show that maps multilinear in k variables are homogeneous of degree k. Give other examples. (iii) If f is C k show that f (e) = (1/k!)Dk f (0) · ek , that is, f may be regarded as an element of S k (E, F) and thus it is C ∞ . Hint: f (0) = 0; inductively applying Taylor’s theorem and replacing at each step h by th, show that f (h) =

1 k 1 D f (0) · hk + k o(tk hk ). k! t

 2.4-6. Let e1 , . . . , en−1 ∈ E be fixed and f : U ⊂ E → F be n times differentiable. Show that the map g : U ⊂ E → F defined by g(u) = Dn−1 f (u) · (e1 , . . . , en−1 ) is differentiable and Dg(u) · e = Dn f (u) · (e, e1 , . . . , en−1 ).  2.4-7. (i) Prove the following refinement of Proposition 2.4.14. If f is C 1 and D1 D2 f (u) exists and is continuous in u, then D2 D1 f (u) exists and these are equal. (ii) The hypothesis in (i) cannot be weakened: show that the function ( xy(x2 −y 2 ) x2 +y 2 , if (x, y) 6= (0, 0); f (x, y) = 0, if (x, y) = (0, 0) is C 1 , has ∂ 2 f /∂x∂y, ∂ 2 f /∂y∂x continuous on R2 \{(0, 0)}, but that ∂ 2 f (0, 0)/∂x∂y 6= ∂ 2 f (0, 0)/∂y∂x.  2.4-8.

For f : U ⊂ E → F, show that the second tangent map is given as follows: T 2 f : (U × E) × (E × E) → (F × F) × (F × F)(u, e1 , e2 , e3 ) 7→ (f (u), Df (u) · e1 , Df (u) · e2 , D2 f (u) · (e1 , e2 ) + Df (u) · e3 ).

 2.4-9. Let f : R2 → R be defined by f (x, y) = 2x2 y/(x4 + y 2 ) if (x, y) 6= (0, 0) and 0 if (x, y) = (0, 0). Show that (i) f is discontinuous at (0, 0), hence is not differentiable at (0, 0); (ii) all directional derivatives exist at (0, 0); that is, f is Gˆateaux differentiable.  2.4-10 (Differentiating sequences). Let fn : U ⊂ E → F be a sequence of C r maps, where E and F are Banach spaces. If {fn } converges pointwise to f : U → F and if {Dj fn }, 0 ≤ j ≤ r, converges locally uniformly to a map g j : U → Ljs (E, F), then show that f is C r , Dj f = g j and {fn } converges locally uniformly to f . Hint: For r = 1 use the mean value inequality and continuity of g 1 to conclude that kf (u + h) − f (u) − g 1 (u) · hk ≤ kf (u + h) − fn (u + h) − [f (u) − fn (u)]k + kfn (u + h) − fn (u) − Dfn (u) · hk + kDfn (u) · h − g 1 (u) · hk ≤ ekhk. For general r use the converse to Taylor’s theorem.

2.4 Properties of the Derivative

 2.4-11 (α Lemma). hence is C ∞ .

101

In the context of Lemma 2.4.21 let α(g) = g ◦ f . Show that α is continuous linear and

 2.4-12. Consider the map Φ : C 1 ([0, 1]) → C 0 ([0, 1]) given by Φ(f )(x) = exp[f 0 (x)]. Show that Φ is C ∞ and compute DΦ.  2.4-13 ( Whitney [1943a]).

Let f : U ⊂ E → F be of class C k+p with Taylor expansion

1 f (b) = f (a) + Df (a) · (b − a) + · · · + Dk f (a) · (b − a)k k! Z 1  (1 − t)k−1 k k + [D f ((1 − t)a + tb) − D f (a)] dt · (b − a)k . (k − 1)! 0 (i) Show that the remainder Rk (a, b) is C k+p for b 6= a and C p for a, b ∈ E. If E = F = R, Rk (a, a) = 0, and lim (|b − a|i Di+p Rk (a, b)) = 0, 1 ≤ i ≤ k. b→a

(For generalizations to Banach spaces, see Tuan and Ang [1979].) (ii) Show that the conclusion in (i) cannot be improved by considering f (x) = |x|k+p+1/2 .  2.4-14 ( Whitney [1943b]). f (x) = −f (−x)).

Let f : R → R be an even (resp., odd) function; that is, f (x) = f (−x) (resp.,

(i) Show that f (x) = g(x2 ) (resp., f (x) = xg(x2 )) for some g. (ii) Show that if f is C 2k (resp., C 2k+1 ) then g is C k Hint: Use the converse to Taylor’s theorem. (iii) Show that (ii) is still true if k = ∞. (iv) Let f (x) = |x|2k+1+1/2 to show that the conclusion in (ii) cannot be sharpened.  2.4-15 (Buchner, Marsden, and Schecter [1983b]). Let E = L4 ([0, 1]) and let ϕ : R → R be a C ∞ function such that ϕ0 (λ) = 1, if −1 ≤ λ ≤ 1 and ϕ0 (λ) = 0, if |λ| ≥ 2. Assume ϕ is monotone increasing with ϕ = −M for λ ≤ −2 and ϕ = M for λ ≥ 2. Define the map h : E → R by h(u) =

1 3

Z

1

ϕ([u(x)]3 ) dx.

0

(i) Show that h is C 3 using the converse to Taylor’s theorem. Hint: Let ψ(λ) = ϕ(λ3 ), write out Taylor’s theorem for r = 3 for ψ(λ), and plug in u(x) for λ. (ii) The formal L2 gradient of h (i.e., the functional derivative δh/δu) is given by ∇h(u) =

1 0 ψ (u), 3

where ψ(λ) = ϕ(λ3 ). Show that ∇h : E → E is C 0 but is not C 1 . Hint: Its derivative would be v 7→ ψ 00 (u)v/3. Let a ∈ [0, 1] be such that ϕ00 (a)/3 6= 0 and let un = a on [0, 1/n], un = 0 elsewhere; vn = n1/4 on [0, 1/n], vn = 0 elsewhere. Show that in L4 ([0, 1]), un → 0, kvn k = 1, ψ 00 (un ) · vn does not converge to 0, but ψ 00 (0) = 0. Using the same method, show h is not C 4 on L4 ([0, 1]). (iii) Show that if q is a positive integer and E = Lq ([0, 1]), then h is C q−1 but is not C q .

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(iv) Let f (u) =

1 2

Z

1

|u(x)|2 dx + h(u)

0

Show that on L4 ([0, 1]), f has a formally non-degenerate critical point at 0 (i.e., D2 f (0) defines an isomorphism of L2 ([0, 1])), yet this critical point is not isolated . Hint: Consider the function un = −1 on [0, 1/n]; 0 on ]1/n, 1]. This exercise is continued in Exercise 5.4-8.  2.4-16. Let E be the space of maps A : R3 → R3 with A(x) → 0 as x → 0 sufficiently rapidly. Let f : E → R and show δf δ f (curl A) = curl . δA δA Hint: Specify whatever smoothness and fall-off hypotheses you need; use A·curl B−B·curl A = div(B×A), the divergence theorem, and the chain rule.  2.4-17. (i) Let E = { B | RB is a vector field on R3 vanishing at ∞ and such that div B = 0 } Rand pair E with itself via hB, B0 i = B(x)·B0 (x) dx. Compute δF/δB, where F is defined by F = (1/2) kBk2 d3x. (ii) Let E = { B | B is a vector field on R3 vanishing at ∞ such that B = ∇ × A for some A } and let F = { A0 | A0 is a vector field on R3 , div A0 = 0 } R with the pairing hB, A0 i = A · A0 d3 x. Show that this pairing is well defined. Compute δF/δB, where F is as in (i) . Why is your answer different?

2.5

The Inverse and Implicit Function Theorems

The inverse and implicit function theorems are pillars of nonlinear analysis and geometry, so we give them special attention in this section. Throughout, E, F, . . . , are assumed to be Banach spaces. In the finitedimensional case these theorems have a long and complex history; the infinite-dimensional version is apparently due to Hildebrandt and Graves [1927]. The Inverse Function Theorem. This theorem states that if the linearization of the equation f (x) = y is uniquely invertible, then locally so is f ; that is, we can uniquely solve f (x) = y for x as a function of y. To formulate the theorem, the following terminology is useful. 2.5.1 Definition. A map f : U ⊂ E → V ⊂ F, where U and V are open subsets of E and F respectively, is a C r diffeomorphism if f is of class C r , is a bijection (i.e., f is one-to-one and onto from U to V ), and f −1 is also of class C r . The example f (x) = x3 shows that a map can be smooth and bijective, but its inverse need not be smooth. A theorem guaranteeing a smooth inverse is the following. 2.5.2 Theorem (Inverse Mapping Theorem). Let f : U ⊂ E → F be of class C r , r ≥ 1, x0 ∈ U , and suppose that Df (x0 ) is a linear isomorphism. Then f is a C r diffeomorphism of some neighborhood of x0 onto some neighborhood of f (x0 ) and, moreover, the derivative of the inverse function is given by Df −1 (y) = [Df (f −1 (y))]−1 for y in this neighborhood of f (x0 ). Although our immediate interest is the finite-dimensional case, for Banach spaces it is good to keep in mind the Banach isomorphism theorem: If T : E → F is linear, bijective, and continuous, then T −1 is continuous. (See Theorem 2.2.19.)

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103

Proof of the Inverse Function Theorem. To prove the theorem, we assemble a few lemmas. First recall the contraction mapping principle from §1.2. 2.5.3 Lemma. Let M be a complete metric space with distance function d : M × M → R. Let F : M → M and assume there is a constant λ, 0 ≤ λ < 1, such that for all x, y ∈ M , d(F (x), F (y)) ≤ λd(x, y). Then F has a unique fixed point x0 ∈ M ; that is, F (x0 ) = x0 . This result is the basis of many important existence theorems in analysis. The other fundamental fixed point theorem in analysis is the Schauder fixed point theorem, which states that a continuous map of a compact convex set (in a Banach space, say) to itself, has a fixed point—not necessarily unique, however. 2.5.4 Lemma.

The set GL(E, F) of linear isomorphisms from E to F is open in L(E, F).

Proof. We can assume E = F. Indeed, if ϕ0 ∈ GL(E, F), the linear isomorphism ψ 7→ ϕ−1 0 ◦ ψ from L(E, F) to L(E, E) is continuous and GL(E, F) is the inverse image of GL(E, E). Let kαk = sup kα(e)k e∈E kek=1

be the operator norm on L(E, F) relative to given norms on E and F. For ϕ ∈ GL(E, E), we need to prove that ψ sufficiently near ϕ is also invertible. We will show that kψ − ϕk < kϕ−1 k−1 implies ψ ∈ GL(E, E). The key is that k · k is an algebra norm. That is, kβ ◦ αk ≤ kβk kαk for α ∈ L(E, E) and β ∈ L(E, E) (see §2.2). Since ψ = ϕ ◦ (I − ϕ−1 ◦ (ϕ − ψ)), ϕ is invertible, and our norm assumption shows that kϕ−1 ◦ (ϕ − ψ)k < 1, it is sufficient to show that I − ξ is invertible whenever kξk < 1. (I is the identity operator.) Consider the following sequence called the Neumann series: ξ0 = I, ξ1 = I + ξ, ξ2 = I + ξ + ξ ◦ ξ, .. . ξn = I + ξ + ξ ◦ ξ + · · · + (ξ ◦ ξ ◦ · · · ◦ ξ). Using the triangle inequality and kξk < 1, we can compare this sequence to the sequence P∞ of real numbers, 1, 1 + kξk, 1 + kξk + kξk2 , . . . , which is a Cauchy sequence since the geometric series n=0 kξkn converges. Because L(E, E) is complete, ξn is a convergent sequence. The limit, say ρ, is the inverse of I − ξ because (I − ξ)ξn = I − (ξ ◦ ξ ◦ · · · ◦ ξ), so letting n → ∞, we get (I − ξ)ρ = I. 

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2.5.5 Lemma.

Let I : GL(E, F) → GL(F, E) be given by ϕ 7→ ϕ−1 . Then I is of class C ∞ and DI(ϕ) · ψ = −ϕ−1 ◦ ψ ◦ ϕ−1 .

(For Dr I, see Supplement 2.5E.) Proof. We may assume GL(E, F) 6= ∅. We claim that I is differentiable and that DI(ϕ) · ψ = −ϕ−1 ◦ ψ ◦ ϕ−1 . If we can show this, then it will follow from Leibniz’ rule that I is of class C ∞ . Indeed DI = B(I, I) where B ∈ L2 (L(F, E); L(L(E, F), L(F, E))) is defined by B(ψ1 , ψ2 )(A) = −ψ1 ◦ A ◦ ψ2 , where ψ1 , ψ2 ∈ L(F, E) and A ∈ L(E, F), which shows inductively that if I is C k then it is C k+1 . We first prove our claim that I is differentiable. Since the map ψ 7→ −ϕ−1 ◦ψ ◦ϕ−1 is linear (ψ ∈ L(E, F)), we must show that kψ −1 − (ϕ−1 − ϕ−1 ◦ ψ ◦ ϕ−1 + ϕ−1 ◦ ϕ ◦ ϕ−1 )k = 0. lim ψ→ϕ kψ − ϕk Note that ψ −1 − (ϕ−1 − ϕ−1 ◦ ψ ◦ ϕ−1 + ϕ−1 ◦ ϕ ◦ ϕ−1 ) = ψ −1 − 2ϕ−1 + ϕ−1 ◦ ψ ◦ ϕ−1 = ψ −1 ◦ (ψ − ϕ) ◦ ϕ−1 ◦ (ψ − ϕ) ◦ ϕ−1 . Again, using kβ ◦ αk ≤ kαk kβk for α ∈ L(E, F) and β ∈ L(F, G), we get kψ −1 ◦ (ψ − ϕ) ◦ ϕ−1 ◦ (ψ − ϕ) ◦ ϕ−1 k ≤ kψ −1 k kψ − ϕk2 kϕ−1 k2 . With this inequality, the limit is clearly zero.



Proof of the Inverse Mapping Theorem. We claim that it is enough to prove it under the simplifying assumptions x0 = 0, f (x0 ) = 0, E = F, and Df (0) is the identity. Indeed, replace f by h(x) = Df (x0 )−1 · [f (x + x0 ) − f (x0 )]. Let g(x) = x − f (x) so Dg(0) = 0. Choose r > 0 so that kxk ≤ r implies kDg(x)k ≤ 1/2, which is possible by continuity of Dg. Thus, by the mean value inequality, kxk ≤ r implies kg(x)k ≤ r/2. Let Bε (0) = { x ∈ E | kxk ≤ ε }. For y ∈ Br/2 (0), let gy (x) = y + g(x). If y ∈ Br/2 (0) and x ∈ Br (0), then kyk ≤ r/2 and kg(x)k ≤ r/2, so kgy (x)k ≤ kyk + kg(x)k ≤ r.

(2.5.1)

For x1 , x2 ∈ Br (0) the mean value inequality gives kgy (x1 ) − gy (x2 )k ≤

kx1 − x2 k . 2

(2.5.2)

This shows that for y in the ball of radius r/2, gy maps the closed ball (a complete metric space) of radius r to itself and is a contraction. Thus by the contraction mapping theorem (Lemma 2.5.3), gy has a unique fixed point x in Br (0). This point x is the unique solution of f (x) = y. Thus f has an inverse f −1 : V0 = Dr/2 (0) → U0 = f −1 (Dr/2 (0)) ⊂ Dr (0). From equation (2.5.2) with y = 0, we have k(x1 − f (x1 )) − (x2 − f (x2 ))k ≤ kx1 − x2 k/2, and so kx1 − x2 k − kf (x1 ) − f (x2 )k ≤

kx1 − x2 k , 2

2.5 The Inverse and Implicit Function Theorems

105

that is, kx1 − x2 k ≤ 2kf (x1 ) − f (x2 )k. Thus, we have kf −1 (y1 ) − f −1 (y2 )k ≤ 2ky1 − y2 k,

(2.5.3)

−1

so f is Lipschitz and hence continuous on U0 . From Lemma 2.5.4 we can choose r small enough so that Df (x)−1 exists for x ∈ Dr (0). Moreover, by continuity, kDf (x)−1 k ≤ M for some M and all x ∈ Dr (0) can be assumed as well. If y1 , y2 ∈ Dr/2 (0), x1 = f −1 (y1 ), and x2 = f −1 (y2 ), then kf −1 (y1 ) − f −1 (y2 ) − Df (x2 )−1 · (y1 − y2 )k

= x1 − x2 − Df (x2 )−1 · [f (x1 ) − f (x2 )] = kDf (x2 )−1 · {Df (x2 ) · (x1 − x2 ) − f (x1 ) + f (x2 )}k ≤ M kf (x1 ) − f (x2 ) − Df (x2 ) · (x1 − x2 )k. This, together with equation (2.5.3), shows that f −1 is differentiable with derivative Df (x)−1 at f (x); that is, D(f −1 ) = I ◦ Df ◦ f −1 on V0 = Dr/2 (0). This formula, the chain rule, and Lemma 2.5.5 show inductively that if f −1 is C k−1 then f −1 is C k for 1 ≤ k ≤ r.  This argument also proves the following: if f : U → V is a C r homeomorphism where U ⊂ E and V ⊂ F are open sets, and Df (u) ∈ GL(E, F) for u ∈ U , then f is a C r diffeomorphism. For a Lipschitz inverse function theorem see Exercise 2.5-11.

Supplement 2.5A The Size of the Neighborhoods in the Inverse Mapping Theorem An analysis of the preceding proof also gives explicit estimates on the size of the ball on which f (x) = y is solvable. Such estimates are sometimes useful in applications. The easiest one to use in examples involves estimates on the second derivative.2 2.5.6 Proposition. Let

Suppose f : U ⊂ E → F is of class C r , r ≥ 2, x0 ∈ U , and Df (x0 ) is an isomorphism. L = kDf (x0 )k

and

M = kDf (x0 )−1 k.

Assume kD2 f (x)k ≤ K

for

kx − x0 k ≤ R

and

BR (x0 ) ⊂ U.

Let N, P, Q and S be defined by 3

N = 8M K, P = min   1 P Q = min , ,P , 2N L M



 1 ,R , 2KM   1 Q S = min , ,Q . 2KM 2L

Then f maps an open set G ⊂ DP (x0 ) diffeomorphically onto DP/2M (y0 ) and f −1 maps an open set H ⊂ DQ (y0 ) diffeomorphically onto DQ/2L (x0 ). Moreover, DQ/2L (x0 ) ⊂ G ⊂ DP (x0 ) and DS/2M (y0 ) ⊂ H ⊂ DQ (y0 ) ⊂ DP/2M (y0 ). See Figure 2.5.1. 2 We

thank M. Buchner for his suggestions concerning this supplement.

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2. Banach Spaces and Differential Calculus

H

G

Figure 2.5.1. Regions for the proof of the inverse mapping theorem

Proof.

We can assume x0 = 0 and f (x0 ) = 0. From Z 1 Df (x) = Df (0) + D(Df (tx)) · x dt 0   Z 1 −1 2 = Df (0) · I + [Df (0)] · D f (tx) · x dt 0

and the fact that k(I + A)−1 k ≤ 1 + kAk + kAk2 + · · · =

1 1 − kAk

for kAk < 1 (see the proof of Lemma 2.5.4), we get kDf (x)−1 k ≤ 2M

if kxk ≤ R and kxk ≤

1 , 2M K

that is, if kxk ≤ P . As in the proof of the inverse function theorem, let gy (x) = [Df (0)]−1 · (y + Df (0)x − f (x)). Write ϕ(x) = Df (0) · x − f (x) Z 1 Z = Dϕ(sx) · x ds = 0

1

(Df (0) · x − Df (sx) · x) ds

0

Z 1Z =− 0

1

D2 f (tsx) · (sx, x) dt ds

0

to obtain gy (x) = [Df (0)−1 ] · (y + ϕ(x)), kϕ(x)k ≤ Kkxk2 if kxk ≤ P , and kgy (x)k ≤ M (kyk + Kkxk2 ). Hence for kyk ≤ P/2M , gy maps BP (0) to BP (0). Similarly we get kgy (x1 ) − gy (x2 )k ≤ kx1 − x2 k/2 from the mean value inequality and the estimate

  Z 1

2

≤ M (Kkxk) ≤ 1 kDgy (x)k = kDf (0)−1 k D f (tx) · x dt

2 0

if kxk ≤ P . Thus, as in the previous proof, f −1 : BP/2M (0) → BP (0) is defined and there exists an open set G ⊂ BP (0) diffeomorphic via f to the open ball DP/2M (0).

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107

Taking the second derivative of the relation f −1 ◦ f = identity on G, we get D2 f −1 (f (x))(Df (x) · u1 , Df (x) · u2 ) + Df −1 (f (x)) · D2 f (x)(u1 , u2 ) = 0 for any u1 , u2 ∈ E. Let vi = Df (x) · ui , i = 1, 2, so that D2 f −1 (f (x)) · (v1 , v2 ) = −Df −1 (f (x)) · D2 f (x)(Df (x)−1 · v1 , Df (x)−1 · v2 ) and hence

2 −1





D f (f (x))(v1 , v2 ) ≤ Df −1 (f (x)) 3 D2 f (x) kv1 k kv2 k ≤ 8M 3 K kv1 k kv2 k since x ∈ G ⊂ DP (0) and on BP (0) we have the inequality kDf (x)−1 k ≤ 2M . Thus on BP/2M (0) the following estimate holds: kD2 f −1 (y)k ≤ 8M 3 K. By the previous argument with f replaced by f −1 , R by P/2M , L by M , and K by N = 8M 3 K, it follows that there is an open set H ⊂ DQ (0), Q = min{1/2KM, Q/2L, Q} such that f −1 : H → DQ/2L (0) is a diffeomorphism. Since f −1 is a diffeomorphism on DQ (0) and H is one of its open subsets, it follows that DQ/2L (0) ⊂ G. Finally, replacing R by Q/2L, we conclude that on the open ball DS/2M (0), where S = min{1/2KM, Q/2L, Q}, f −1 is a diffeomorphism. Therefore DS/2M (0) ⊂ H. 

Implicit Function Theorem. In the study of manifolds and submanifolds, the implicit function theorem is of central importance. Like the inverse function theorem, this is a local, nonlinear extension of a simple linear algebra result. Namely, if we have a linear system of equations w = Ax + By, where w ∈ Rn , x ∈ Rm , y ∈ Rn , A is an n × m matrix and B is an n × n matrix, then one can solve for y as a function of w and x if B is invertible. Of course the solution is y = B −1 (w − Ax). Another way to look at it, which is useful for the nonlinear case is to consider the augmented system x=x w = Ax + By for (x, y) ∈ Rm × Rn . If B is invertible, so is the matrix for this system, namely   I 0 A B whose inverse gives the solution    x I = y −B −1 A

0 B −1

  x . w

This trick changes the problem of solving a nonsquare linear system to the problem of solving a square linear system. Likewise in the nonlinear case, an analogous trick enables us to reduce the implicit function theorem to the inverse function theorem. 2.5.7 Theorem (Implicit Function Theorem). Let U ⊂ E, V ⊂ F be open and f : U × V → G be C r , r ≥ 1. For some x0 ∈ U, y0 ∈ V assume the partial derivative in the second argument D2 f (x0 , y0 ) : F → G, is an isomorphism. Then there are neighborhoods U0 of x0 and W0 of f (x0 , y0 ) and a unique C r map g : U0 × W0 → V such that for all (x, w) ∈ U0 × W0 , f (x, g(x, w)) = w.

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Proof.

Define the map Φ:U ×V →E×G

by (x, y) 7→ (x, f (x, y)). Then DΦ(x0 , y0 ) is given by    I 0 x1 DΦ(x0 , y0 ) · (x1 , y1 ) = D1 f (x0 , y0 ) D2 f (x0 , y0 ) y1 which is an isomorphism of E × F with E × G. Thus, Φ has a unique C r local inverse, say Φ−1 : U0 × W0 → U × V , (x, w) 7→ (x, g(x, w)). The g so defined is the desired map.  Applying the chain rule to the relation f (x, g(x, w)) = w, one can compute the derivatives of g: D1 g(x, w) = −[D2 f (x, g(x, w))]−1 ◦ D1 f (x, g(x, w)), D2 g(x, w) = [D2 f (x, g(x, w))]−1 . 2.5.8 Corollary. Let U ⊂ E be open and f : U → F be C r , r ≥ 1. Suppose Df (x0 ) is surjective and ker Df (x0 ) has a closed complement. Then f (U ) contains a neighborhood of f (x0 ). Proof. Let E1 = kerDf (x0 ) and E = E1 ⊕ E2 . Then D2 f (x0 ) : E2 → F is an isomorphism, so the hypotheses of Theorem 2.5.7 are satisfied and thus f (U ) contains W0 provided by that theorem.  Local Surjectivity Theorem. Since in finite-dimensional spaces every subspace splits, the foregoing corollary implies that if f : U ⊂ Rn → Rm , n ≥ m, and the Jacobian of f at every point of U has rank m, then f is an open mapping. This statement generalizes directly to Banach spaces, but it is not a consequence of the implicit function theorem anymore, since not every subspace is split. This result goes back to Graves [1950]. The proof given in Supplement 2.5B follows Luenberger [1969]. 2.5.9 Theorem (Local Surjectivity Theorem). If f : U ⊂ E → F is C 1 and Df (u0 ) is onto for some u0 ∈ U , then f is locally onto; that is, there exist open neighborhoods U1 of u0 and V1 of f (u0 ) such that f |U1 : U1 → V1 is onto. In particular, if Df (u) is onto for all u ∈ U , then f is an open mapping.

Supplement 2.5B Proof of the Local Surjectivity Theorem Proof. Recall from §2.1 that E/ ker Df (u0 ) = E0 is a Banach space with norm k[x]k = inf{ kx + uk | u ∈ kerDf (u0 ) }, where [x] is the equivalence class of x. To solve f (x) = y we set up an iteration scheme in E0 and E simultaneously. Since Df (u0 ) induces an isomorphism T : E0 → F, it follows that T −1 ∈ L(F, E0 ) exists by the Banach isomorphism theorem. Let x = u0 + h and write f (x) = y as T −1 (y − f (u0 + h)) = 0. To solve this equation, define a sequence Ln ∈ E/ ker Df (u0 ) (so the element Ln is a coset of ker Df (u0 )) and hn ∈ Ln ⊂ E inductively by L0 = ker Df (u0 ), h0 ∈ L0 small, and Ln = Ln−1 + T −1 (y − f (u0 + hn−1 )),

(2.5.4)

khn − hn−1 k ≤ 2kLn − Ln−1 k.

(2.5.5)

and selecting hn ∈ Ln such that The latter is possible since kLn − Ln−1 k = inf{ kh − hn−1 k | h ∈ Ln }.

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109

Since hn−1 ∈ Ln−1 , Ln−1 = T −1 (Df (u0 ) · hn−1 ), so Ln = T −1 (y − f (u0 + hn−1 ) + Df (u0 ) · hn−1 ). Subtracting this from the expression for Ln−1 gives Ln − Ln−1 = −T −1 (f (u0 + hn−1 ) − f (u0 + hn−2 ) − Df (u0 ) · (hn−1 − hn−2 ))  Z 1 (Df (u0 + thn−1 + (1 − t)hn−2 ) − Df (u0 )) dt · (hn−1 − hn−2 ) . = −T −1 0

Using Proposition 2.4.7, for ε > 0 given, there is a convex neighborhood U of u0 such that kDf (u) − Df (u0 )k < ε for u ∈ U , since f is C 1 . Assume inductively that u0 + hn−1 ∈ U and u0 + hn−2 ∈ U . Then u0 + thn−1 + (1 − t)hn−2 = (1 − t)(u0 + hn−2 ) + t(u0 + hn−1 ) ∈ U for t ∈ [0, 1] and hence kLn − Ln−1 k ≤ εkT −1 k khn−1 − hn−2 k.

(2.5.6)

By equation (2.5.5), khn − hn−1 k ≤ 2kLn − Ln−1 k ≤ 2εkT −1 k khn−1 − hn−2 k. Thus, if ε is small, khn − hn−1 k ≤

1 khn−1 − hn−2 k. 2

Starting with h0 small and kh1 − h0 k < (1/2)kh0 k, u0 + hn remain inductively in U since khn k ≤ kh0 k + kh1 − h0 k + kh2 − h1 k + · · · + khn − hn−1 k   1 1 ≤ 1 + + · · · + n−1 kh0 k ≤ 2kh0 k. 2 2 In addition, khn − hn−1 k <

1 kh0 k 2n

so that

1 kh0 k 2n−1 for all k = 1, 2, 3, . . .. It follows that hn is a Cauchy sequence, so it converges to some point, say h. Correspondingly, Ln converges to L and h ∈ L. Thus from equation (2.5.4), 0 = T −1 (y − f (u0 + h)) and so y = f (u0 + h).  khn+k − hn k <

The local surjectivity theorem shows that for y near y0 = f (u0 ), f (x) = y has a solution. If there is a solution g(y) = x which is C 1 , then Df (x0 ) ◦ Dg(y0 ) = I and so range Dg(y0 ) is an algebraic complement to ker Df (x0 ). It follows that if range Dg(y0 ) is closed, then ker Df (x0 ) is split. In many applications to nonlinear partial differential equations, methods of functional analysis and elliptic operators can be used to show that ker Df (x0 ) does split, even in Banach spaces. Such a splitting theorem is called the Fredholm alternative. For illustrations of this idea in geometry and relativity, see Fischer and Marsden [1975, 1979], and in elasticity, see Chapter 6 of Marsden and Hughes [1983]. For such applications, Corollary 2.5.8 often suffices.

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Local Injectivity Theorem.

The locally injective counterpart of this theorem is the following.

2.5.10 Theorem (Local Injectivity Theorem). Let f : U ⊂ E → F be a C 1 map, Df (u0 )(E) be closed in F, and Df (u0 ) ∈ GL(E, Df (u0 )(E)). Then there exists a neighborhood V of u0 , V ⊂ U , on which f is injective. The inverse f −1 : f (V ) → U is Lipschitz continuous. Proof. Since (Df (u0 ))−1 ∈ L(Df (u0 )(E), E), there is a constant M > 0 such that kDf (u0 ) · ek ≥ M kek for all e ∈ E. By continuity of Df , there exists r > 0 such that kDf (u) − Df (u0 )k < M/2 whenever ku − u0 k < 3r. For e1 , e2 ∈ Dr (u0 ), the identity  Z 1 (Df (e2 + t(e1 − e2 )) − D(u0 )) dt · (e1 − e2 ) f (e1 ) − f (e2 ) − Df (u0 ) · (e1 − e2 ) = 0

(see Proposition 2.4.7) implies kf (e1 ) − f (e2 ) − Df (u0 ) · (e1 − e2 )k ≤ sup kDf (e2 + t(e1 − e2 )) − Df (u0 )k ke1 − e2 k t∈[0,1]

≤

M ke1 − e2 k 2

since ku0 − e1 − t(e2 − e1 )k < 3r. Thus M ke1 − e2 k ≤ kDf (u0 ) · (e1 − e2 )k ≤ kf (e1 ) − f (e2 )k +

M ke1 − e2 k; 2

that is, M ke1 − e2 k ≤ kf (e1 ) − f (e2 )k, 2 which proves that f is injective on Dr (u0 ) and that f −1 : f (Dr (u0 )) → U is Lipschitz continuous.



Notice that this proof is done by direct estimates, and not by invoking the inverse or implicit function theorem. If, however, the range space Df (u0 )(E) splits, one could alternatively prove results like this by composing f with the projection onto this range and applying the inverse function theorem to the composition. In the following paragraphs on local immersions and submersions, we examine this point of view in detail. Application to Differential Equations. We now give an example of the use of the implicit function theorem to prove an existence theorem for differential equations. For this and related examples, we choose the spaces to be infinite dimensional. In fact, E, F, G, · · · will be suitable spaces of functions. The map f will often be a nonlinear differential operator. The linear map Df (x0 ) is called the linearization of f about x0 . (Phrases like “first variation,” “first-order deformation,” and so forth are also used.) 2.5.11 Example.

Let E be the space of all C 1 -functions f : [0, 1] → R with the norm df (x) kf k1 = sup |f (x)| + sup dx x∈[0,1] x∈[0,1]

and F the space of all C 0 -functions with the norm kf k0 = supx∈[0,1] |f (x)|. These are Banach spaces (see Exercise 2.1-3). Let Φ : E → F be defined by Φ(f ) = df /dx + f 3 . It is easy to check that Φ is C ∞ and DΦ(0) = d/dx : E → F. Clearly DΦ(0) is surjective (fundamental theorem of calculus). Also ker DΦ(0) consists of E1 = all constant functions. This is complemented—a closed complement consists of functions with zero integral. Thus, Corollary 2.5.8 yields the following: There is an ε > 0 such that if g : [0, 1] → R is a continuous function with |g(x)| < ε, then there is a C 1 function f : [0, 1] → R such that df + f 3 (x) = g(x).  dx

2.5 The Inverse and Implicit Function Theorems

111

Supplement 2.5C An Application of the Inverse Function Theorem to a Nonlinear Partial Differential Equation Let Ω ⊂ Rn be a bounded open set with smooth boundary. Consider the problem ∇ 2 ϕ + ϕ3 = f

in Ω,

ϕ + ϕ7 = g

on ∂Ω

for given f and g. (The powers of 3 and 7 have nothing special about them and are just for illustrative purposes). We claim that for f and g small, this problem has a unique small solution. For partial differential equations of this sort one can use the Sobolev spaces H s (Ω, R) consisting of maps ϕ : Ω → R whose first s distributional derivatives lie in L2 . (One uses Fourier transforms to define this space if s is not an integer.) In the Sobolev spaces E = H s (Ω, R), F = H s−2 (Ω, R) × H s−1/2 (∂Ω, R), if s > n/2 the map Φ : E → F,

ϕ 7→ (∇2 ϕ + ϕ3 , (ϕ + ϕ7 )|∂Ω)

is C ∞ (use Supplement 2.4B) and the linear operator DΦ(0) · ϕ = (∇2 ϕ, ϕ|∂Ω) is an isomorphism. The fact that DΦ(0) is an isomorphism is a result on the solvability of the Dirichlet problem from the theory of elliptic linear partial differential equations. See, for example, Friedman [1969]. (In the C k spaces, DΦ(0) is not an isomorphism.) The result claimed above now follows from the inverse function theorem.

Local Immersions and Submersions. The following series of consequences of the inverse function theorem are important technical tools in the study of manifolds. The first two results give, roughly speaking, sufficient conditions to “straighten out” the range (respectively, the domain) of f in a neighborhood of a point, thus making f look like an inclusion (respectively, a projection). 2.5.12 Theorem (Local Immersion Theorem). Let f : U ⊂ E → F be of class C r , r ≥ 1, u0 ∈ U and suppose that Df (u0 ) is one-to-one and has closed split image F1 with closed complement F2 . (If E = Rm and F = Rn , assume only that Df (u0 ) has trivial kernel.) Then there are two open sets U 0 ⊂ F and V ⊂ E ⊕ F2 and a C r diffeomorphism ϕ : U 0 → V satisfying f (u0 ) ∈ U 0 , u0 ∈ V ∩ (E × {0}) ⊂ U , and (ϕ ◦ f )(e) = (e, 0) for all e ∈ V ∩ (E × {0}) ⊂ U ⊂ E. In addition, the Banach space isomorphism Dϕ(u0 ) : F = F1 ⊕ F2 → E ⊕ F2 is the identity on F2 for any u0 ∈ U 0 and Dϕ(f (u0 )) is a block diagonal operator. The intuition for E = F1 = R2 , F2 = R (i.e. m = 2, n = 3) is given in Figure 2.5.2. The function ϕ flattens out the image of f . Notice that this is intuitively correct; we expect the range of f to be an m-dimensional “surface” so it should be possible to flatten it to a piece of Rm . Note that the range of a linear map of rank m is a linear subspace of dimension exactly m, so this result expresses, in a sense, a generalization of the linear case. Also note that Theorem 2.5.10, the local injectivity theorem, follows from the more restrictive hypotheses of Theorem 2.5.12. Proof. Now

Define g : U × F2 ⊂ E ⊕ F2 → F = F1 ⊕ F2 by g(u, v) = f (u) + (0, v) and note that g(u, 0) = f (u). Dg(u0 , 0) = (Df (u0 ), IF2 ) ∈ GL(E ⊕ F2 , F)

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f

f(u0)

V

Figure 2.5.2. The local immersion theorem

by the Banach isomorphism theorem. Here, IF2 denotes the identity mapping of F2 and for A ∈ L(E, F) and B ∈ L(E0 , F0 ), the element (A, B) ∈ L(E ⊕ E0 , F ⊕ F0 ) is defined by (A, B)(e, e0 ) := (Ae, Be0 ). By the inverse function theorem there exist open sets U 0 and V and a C r diffeomorphism ϕ : U 0 → V such that (u0 , 0) ∈ V ⊂ U × F2 ⊂ E ⊕ F2 , g(u0 , 0) = f (u0 ) ∈ U 0 ⊂ F, and ϕ−1 = g|V . Hence for (e, 0) ∈ V , (ϕ ◦ f )(e) = (ϕ ◦ g)(e, 0) = (e, 0). Writing f (u) = (f1 (u), f2 (u)) ∈ F1 ⊕ F2 , it follows that g(u, v) = (f1 (u), f2 (u) + v) and hence that   Df1 (u) 0 Dg(u, v) = : E ⊕ F2 → F1 ⊕ F2 Df2 (u) IF2 for all (u, v) ∈ V . Thus Dg(u0 , 0), regarded as a linear continuous isomorphism from E ⊕ F2 to F1 ⊕ F2 , is a block diagonal operator because Df (u0 )(E) = F1 and hence Df1 (u0 ) = Df (u0 ) and Df2 (u0 ) = 0. Therefore its inverse, Dϕ(f (u0 )), will be also block diagonal. In addition, Dg(u, v)|{0} × F2 is the identity on F2 for all (u, v) ∈ V , whence Dϕ(u0 )|{0} × F2 is also the identity for all u0 ∈ U .  2.5.13 Theorem (Local Submersion Theorem). Let f : U ⊂ E → F be of class C r , r ≥ 1, u0 ∈ U and suppose that Df (u0 ) is surjective and has split kernel E2 with closed complement E1 . (If E = Rm and F = Rn , assume only that rank(Df (u0 )) = n.) Then there are two open sets U 0 ⊂ U ⊂ E = E1 ⊕ E2 and V ⊂ F ⊕ E2 and a C r diffeomorphism ψ : V → U 0 satisfying u0 = (u01 , u02 ) ∈ U 0 , (f (u0 ), u02 ) ∈ V , and (f ◦ ψ)(u, v) = u

and

ψ(u, v) = (ψ1 (u, v), v)

for all (u, v) ∈ V , where ψ1 : V → E1 is a C r map. In addition, the Banach space isomorphism Dψ(f (u0 ), u02 ) : F ⊕ E2 → E = E1 ⊕ E2 is a block diagonal operator and Dψ(u, v)|F × {0} : F × {0} → E1 × {0} is a Banach space isomorphism for all (u, v) ∈ V . The intuition for the special case E1 = E2 = F = R is given in Figure 2.5.3, which should be compared to Figure 2.5.2. Note also that this theorem implies the results of Theorem 2.5.9, the local surjectivity theorem, but the hypotheses are more stringent. Proof.

By the Banach isomorphism theorem (§2.2), D1 f (u0 ) = Df (u0 )|E1 ∈ GL(E1 , F). Define the map g : U ⊂ E1 ⊕ E2 → F ⊕ E2

2.5 The Inverse and Implicit Function Theorems

113

f V F

F

Figure 2.5.3. The local submersion theorem

by g(u1 , u2 ) = (f (u1 , u2 ), u2 ). Therefore  Dg(u) · (e1 , e2 ) =

D1 f (u) D2 f (u) 0 IE2



e1 e2



for all u = (u1 , u2 ) ∈ U , e1 ∈ E1 , e2 ∈ E2 . By hypothesis E2 = ker Df (u0 ) and hence D2 f (u0 ) = Df (u0 )|E2 = 0. Therefore Dg(u0 ) is a block diagonal operator. Thus Dg(u0 ) ∈ GL(E, F ⊕ E2 ). By the inverse function theorem, there are open sets U 0 and V and a C r diffeomorphism ψ : V → U 0 such that u0 ∈ U 0 ⊂ U ⊂ E, g(u0 ) ∈ V ⊂ F ⊕ E2 , and ψ −1 = g|U 0 . Hence if (u, v) ∈ V , then (u, v) = (g ◦ ψ)(u, v) = (f (ψ(u, v)), ψ2 (u, v)), where ψ = ψ1 × ψ2 . This shows that ψ2 (u, v) = v and (f ◦ ψ)(u, v) = u. Since Dg(u0 ) is block diagonal, so is its inverse Dψ(f (u0 ), u02 ), where u0 = (u01 , u02 ). Moreover, since Dg(u0 ) is a Banach space isomorphism for all u0 ∈ U 0 , its expression as a matrix operator above shows that D1 f (u0 ) : E1 → F is also a Banach space isomorphism, that is, Dg(u0 )|E1 × {0} : E1 × {0} → F × {0} is a Banach space isomorphism. Consequently, Dψ(u, v)|F × {0} : F × {0} → E1 × {0} is a Banach space isomorphism for all (u, v) ∈ V .  Local Representation and Rank Theorems. on the local structure of maps.

We now give two results that extend the above theorems

2.5.14 Theorem. (Local Representation Theorem) Let f : U ⊂ E → F be of class C r , r ≥ 1, u0 ∈ U and suppose that Df (u0 ) has closed split image F1 with closed complement F2 and split kernel E2 with closed complement E1 . (If E = Rm and F = Rn , assume only that rank(Df (u0 )) = k, k ≤ n, k ≤ m, so that F2 = Rn−k , F1 = Rk , E1 = Rk , E2 = Rm−k .) Then there are two open sets U 0 ⊂ U ⊂ E = E1 ⊕ E2 and V ⊂ F1 ⊕ E2 and a C r diffeomorphism ψ : V → U 0 satisfying u0 = (u01 , u02 ) ∈ U 0 , (f (u0 ), u02 ) ∈ V , and (f ◦ ψ)(u, v) = (u, η(u, v))

for all

(u, v) ∈ V,

where η : V → F2 is a C r map satisfying Dη(ψ −1 (u0 )) = 0. Let f = f1 × f2 , where fi : U → Fi , i = 1, 2. Using this notation, the Banach space isomorphism Dψ(f1 (u0 ), u02 ) : F1 ⊕ E2 → E = E1 ⊕ E2 is a block diagonal operator and Dψ(u, v)|F1 × {0} : F1 × {0} → E1 × {0} is a Banach space isomorphism for all (u, v) ∈ V . Proof. Note that f1 satisfies the conditions of Theorem 2.5.13 and thus there exists a C r diffeomorphism ψ : V ⊂ F1 ⊕ E2 → U 0 ⊂ U ⊂ E such that the composition f1 ◦ ψ is given by (f1 ◦ ψ)(u, v) = u and all the other conclusions of the theorem about ψ and its derivatives hold. Define η = f2 ◦ ψ; then (f ◦ ψ)(u, v) = (u, η(u, v)). Since Df2 (u0 ) = 0, the chain rule gives Dη(ψ −1 (u0 )) = 0. 

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To use Theorem 2.5.12 (or Theorem 2.5.13) in finite dimensions, we must have the rank of Df (u0 ) equal to the dimension of its domain space (or the range space). However, we can also use the inverse function theorem to tell us that if Df (u) has constant rank k for u in a neighborhood of u0 , then we can straighten out the domain of f with some invertible function ψ such that f ◦ ψ depends only on k variables. Then we can apply the local immersion theorem (Theorem 2.5.12) to straighten out the range. This is the essence of the following theorem. Roughly speaking, in finite dimensions, the rank theorem says that if Df has constant rank k on an open set in Rm , then m − k variables are redundant and can be eliminated. As a simple example, if f : R2 → R2 is defined by setting f (x, y) = (x − y, −2x + 2y), then Df has everywhere rank 1, and indeed, we can express f using just one variable, namely, let ψ(x, y) = (x + y, y) so that (f ◦ ψ)(x, y) = (x, −2x), which depends only on x. But we can do even better by letting ϕ(x, y) = (−x − y, 2x + y) and observing that (ϕ ◦ f ◦ ψ)(x, y) = (x, 0). Note that both ψ and ϕ are in this case linear isomorphisms. 2.5.15 Theorem (Rank Theorem). Let f : U ⊂ E → F be of class C r , r ≥ 1, u0 ∈ U and suppose that Df (u0 ) has closed split image F1 with closed complement F2 and split kernel E2 with closed complement E1 . In addition, assume that for all u in a neighborhood of u0 ∈ U , Df (u)(E) is a closed subspace of F and Df (u)|E1 : E1 → Df (u)(E) is a Banach space isomorphism. (If E = Rm and F = Rn , assume only that rank(Df (u)) = k, k ≤ n, k ≤ m, for all u in a neighborhood of u0 . In this case, F2 = Rn−k , F1 = Rk , E1 = Rk , E2 = Rm−k .) Then there are open sets U1 ⊂ F1 ⊕ E2 , U2 ⊂ U ⊂ E, V1 ⊂ F, and V2 ⊂ F and two C r diffeomorphism ϕ : V1 → V2 and ψ : U1 → U2 satisfying u0 = (u01 , u02 ) ∈ U2 ⊂ U ⊂ E1 ⊕ E2 , f (u0 ) ∈ V1 , and (ϕ ◦ f ◦ ψ)(x, e) = (x, 0) for all (x, e) ∈ U1 . Let f = f1 × f2 , where fi : U → Fi , i = 1, 2. Using this notation, the Banach space isomorphism Dψ(f1 (u0 ), u02 ) : F1 ⊕ E2 → E = E1 ⊕ E2 is a block diagonal operator and Dψ(x, y)|F1 × {0} : F1 × {0} → E1 × {0} is a Banach space isomorphism for all (x, y) ∈ U1 . The Banach space isomorphism Dϕ(v) : F1 ⊕ F2 → F1 ⊕ F2 is the identity on F2 for any v ∈ V1 and Dϕ(f (u0 )) is a block diagonal operator. The intuition is given by Figure 2.5.4 for E = R2 , F = R2 , and k = 1. Proof. By the local representation theorem there is a C r diffeomorphism ψ : U1 ⊂ F1 ⊕E2 → U2 ⊂ U ⊂ E, U1 , U2 open, u0 ∈ U2 , such that f¯(x, y) = (f ◦ ψ)(x, y) = (x, η(x, y)), where η : U1 → F2 is of class C r and satisfies Dη(ψ −1 (u0 )) = 0. The other conclusions about ψ also hold. Shrink U1 (and hence also U2 ), if necessary, in order to insure that for all (x, y) ∈ U1 the space Df¯(x, y)(F1 ⊕ E2 ) is closed in F and the linear map Df¯(x, y)|F1 × {0} : F1 × {0} → Df¯(x, y)(F1 ⊕ E2 ) is a Banach space isomorphism; this is guaranteed by the hypotheses of the theorem as we shall prove now. Since Dψ(x, y) : F1 ⊕ E2 → E1 ⊕ E2 = E is a Banach space isomorphism and Df¯(x, y) = Df (ψ(x, y)) ◦ Dψ(x, y), it follows that Df¯(x, y)(F1 ⊕ E2 ) = Df (ψ(x, y))(E). Since the restriction Dψ(x, y)|F1 ⊕ {0} : F1 ⊕ {0} → E1 ⊕ {0} is also a Banach space isomorphism, the hypothesis of the theorem, namely that for all u is a neighborhood of u0 the range Df (u)(E) is a closed subspace of F and Df (u)|E1 : E1 → Df (u)(E) is a Banach space isomorphism, implies that Df¯(x, y)|F1 ⊕ {0} : F1 ⊕ {0} → Df¯(x, y)(F1 ⊕ E2 ) is a Banach space isomorphism for all (x, y) in the inverse image by ψ of this neighborhood of u0 in U . Note that this implies that Df¯(x, y)(F1 ⊕{0}) = Df¯(x, y)(F1 ⊕E2 ). (In the finite dimensional case this proof is simpler: the expression f¯(x, y) = (x, η(x, y)) immediately implies that Df¯(x, y)|F1 ⊕ {0} is injective and, by hypothesis, dim F1 = dim(Df¯(x, y)(F1 ⊕ E2 )). If P1 : F = F1 ⊕ F2 → F1 is the projection, then P1 |Df¯(x, y)(F1 ⊕ E2 ) : Df¯(x, y)(F1 ⊕ E2 ) → F1 is the inverse of Df¯(x, y)|F1 × {0} : F1 × {0} → Df¯(x, y)(F1 × {0}) = Df¯(x, y)(F1 × E2 ) ⊂ F = F1 ⊕ F2 . Indeed, since Df¯(x, y)·(w, e) = (w, Dη(x, y)·(w, e)), for w ∈ F1 , e ∈ E2 , it follows that (P1 ◦Df¯(x, y))(w, 0) = (w, 0). Therefore, we will also have Df¯(x, y) ◦ P1 |Df¯(x, y)(F1 ⊕ E2 ) = identity on Df¯(x, y)(F1 ⊕ E2 ).

2.5 The Inverse and Implicit Function Theorems

115

f

Figure 2.5.4. The rank theorem

Let (w, Dη(x, y) · (w, e)) ∈ Df¯(x, y)(F1 ⊕ E2 ). Since (Df¯(x, y) ◦ P1 )(w, Dη(x, y) · (w, e)) = Df¯(x, y) · (w, 0) = (w, Dη(x, y) · (w, 0)) = (w, D1 η(x, y) · w), it follows that D2 η(x, y) · e = 0 for all e ∈ E2 , that is, D2 η(x, y) = 0. However, since D2 f¯(x, y) · e = (0, D2 η(x, y) · e), this in turn implies that D2 f¯(x, y) = 0 ∈ F, that is, f¯(x, y) does not depend on the variable y ∈ E2 . Let P10 : F1 ⊕ E2 → F1 be the projection. Define the map f˜ : P10 (U1 ) ⊂ F1 → F by f˜(x) = f¯(x, y) = (f ◦ ψ)(x, y). This function f˜ satisfies the conditions of Theorem 2.5.12 at P10 (ψ −1 (u0 )). Indeed, for any x ∈ P10 (U1 ), the map Df˜(x) = Df¯(x, y)|F1 × {0} : F1 → Df¯(x, y)(F1 ⊕ E2 ) = Df˜(x)(F1 ) is a Banach space isomorphism and Df˜(P10 (ψ −1 (u0 )))(F1 ) = Df¯(ψ −1 (u0 ))(F1 ⊕ E2 ) = Df¯(ψ −1 (u0 ))(Dψ −1 (u0 )(E)) = D(f¯ ◦ ψ −1 )(u0 )(E) = Df (u0 )(E) is closed and has closed split image in F. Thus, by Theorem 2.5.12, there is a C r diffeomorphism ϕ : V1 → V2 , where V1 , V2 ⊂ F are open and f (u0 ) ∈ V1 , u0 ∈ V2 ∩ (F1 × {0}) ⊂ P10 (U1 ) such that (ϕ ◦ f˜)(x) = (x, 0) for all x ∈ V2 ∩(F1 ×{0}). This shows that (ϕ◦f ◦ψ)(x, y) = (x, 0) for all (x, y) ∈ U1 ∩ψ −1 (f −1 (V1 )) ⊂ U1 ⊂ F1 ⊕E2 , so shrinking U1 further if necessary gives the result.  2.5.16 Example (Functional Dependence). Let U ⊂ Rn be an open set and let the functions f1 , . . . , fn : U → R be smooth. The functions f1 , . . . , fn are said to be functionally dependent at x0 ∈ U if there

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is a neighborhood V of the point (f1 (x0 ), . . . , fn (x0 )) ∈ Rn and a smooth function F : V → R such that DF 6= 0 on a neighborhood of (f1 (x0 ), . . . , fn (x0 )), and F (f1 (x), . . . , fn (x)) = 0 for all x in some neighborhood of x0 . Show: (i) If f = (f1 , . . . , fn ) and f1 , . . . , fn are functionally dependent at x0 , then the determinant of Df , denoted ∂(f1 , . . . , fn ) , Jf = ∂(x1 , . . . , x1 ) vanishes at x0 . (ii) If ∂(f1 , . . . , fn−1 ) 6= 0 and ∂(x1 , . . . , xn−1 )

∂(f1 , . . . , fn ) =0 ∂(x1 , . . . , xn )

on a neighborhood of x0 , then f1 , . . . , fn are functionally dependent, and fn = G(f1 , . . . , fn−1 ) for some G. Solution.

(i) We have F ◦ f = 0, so DF (f (x)) ◦ Df (x) = 0.

Now if Jf (x0 ) 6= 0, Df (x) would be invertible in a neighborhood of x0 , implying DF (f (x)) = 0. By the inverse function theorem, this implies DF (y) = 0 on a whole neighborhood of f (x0 ). (ii) The conditions of (ii) imply that Df has rank n − 1. Hence by the rank theorem, there are mappings ϕ and ψ such that (ϕ ◦ f ◦ ψ)(x1 , . . . , xn ) = (x1 , . . . , xn−1 , 0). Let F be the last component of ϕ . Then F (f1 , . . . , fn ) = 0. Since ϕ is invertible, DF 6= 0. It follows from the implicit function theorem that we can locally solve F (f1 , . . . , fn ) = 0 for fn = G(f1 , . . . , fn−1 ), provided we can show ∆ = ∂F/∂yn 6= 0. As we saw before, DF (f (x)) ◦ Df (x) = 0, or, in components with y = f (x), ∂f1    ∂x1 ∂F ∂F   .. ··· . ∂y1 ∂yn   ∂f n ∂x1 

 ∂f1 ∂xn  ..   .  = (0, 0, . . . , 0). ∂f 

···

n

···

∂xn

If ∂F/∂yn = 0, we would have  ∂f 1    ∂x1  . ∂F ∂F  . ,..., . ∂y1 ∂yn−1   ∂f n−1 ∂x1

···

···

∂f1  ∂xn−1  ..   .  = (0, 0, . . . , 0) ∂fn−1  ∂xn−1

i.e., 

∂F ∂F ,..., ∂y1 ∂yn−1

 = (0, 0, . . . , 0)

2.5 The Inverse and Implicit Function Theorems

117

since the square matrix is invertible by the assumption that ∂(f1 , . . . , fn−1 ) 6= 0. ∂(x1 , . . . , xn−1 ) This implies DF = 0, which is not true. Hence ∂F/∂yn 6= 0, and we have the desired result.



Note the analogy between linear dependence and functional dependence, where rank or determinant conditions are replaced by the analogous conditions on the Jacobian matrix.

Supplement 2.5D The Hadamard–Levy Theorem This supplement gives sufficient conditions which together with the hypotheses of the inverse function theorem guarantee that a C k map f between Banach spaces is a global diffeomorphism. To get a feel for these supplementary conditions, consider a C k function f : R → R, k ≥ 1, satisfying 1/|f 0 (x)| < M for all x ∈ R. Then f is a local diffeomorphism at every point of R and thus is an open map. In particular, f (R) is an open interval ]a, b[. The condition |f 0 (x)| > 1/M implies that f is either strictly increasing or strictly decreasing. Let us assume that f is strictly increasing. If b < +∞, then the line y = b is a horizontal asymptote of the graph of f and therefore we should have limx→∞ f 0 (x) = 0 contradicting |f 0 (x)| > 1/M . One similarly shows that a = −∞ and the same proof works if f 0 (x) < −1/M . The theorem below generalizes this result to the case of Banach spaces. 2.5.17 Theorem (The Hadamard–Levy Theorem). Let f : E → F be a C k map of Banach spaces, k ≥ 1. If Df (x) is an isomorphism of E with F for every x ∈ E and if there is a constant M > 0 such that kDf (x)−1 k < M for all x ∈ E, then f is a diffeomorphism. The key to the proof of the theorem consists of a homotopy lifting argument. If X is a topological space, a continuous map ϕ : X → F is said to lift to E through f , if there is a continuous map ψ : X → E satisfying f ◦ ψ = ϕ. 2.5.18 Lemma. Let X be a connected topological space, ϕ : X → F a continuous map and let f : E → F be a C 1 map with Df (e) an isomorphism for every e ∈ E. Fix u0 ∈ E, v0 ∈ F, and x0 ∈ X satisfying f (u0 ) = v0 and ϕ(x0 ) = v0 . Then if a lift ψ of ϕ through f with ψ(x0 ) = u0 exists, it is unique. Proof.

Let ψ 0 be another lift with ψ 0 (x0 ) = u0 and define the sets X1 = { x ∈ X | ψ(x) = ψ 0 (x) } and X2 = { x ∈ X | ψ(x) 6= ψ 0 (x) },

so that X = X1 ∪ X2 and X1 ∩ X2 = ∅. We shall prove that both X1 , X2 are open. Since x0 ∈ X1 , connectedness of X implies X2 = ∅ and the lemma will be proved. If x ∈ X1 , let U be an open neighborhood of ψ(x) = ψ 0 (x) on which f is a diffeomorphism. Then −1 ψ (U ) ∩ ψ 0−1 (U ) is an open neighborhood of x contained in X1 . If x ∈ X2 , let U (resp., U 0 ) be an open neighborhood of the point ψ(x) (resp. of ψ 0 (x)) on which f is a diffeomorphism and such that U ∩ U 0 = ∅. Then the set ψ −1 (U ) ∩ ψ 0−1 (U 0 ) is an open neighborhood of x contained in X2 .  A path γ : [0, 1] → G, where G is a Banach space, is called C 1 if γ|]0, 1[ is uniformly C 1 (that is, it is C 1 and its first derivative is uniformly continuous on ]0, 1[ ) and the extension by continuity of γ 0 to [0, 1] has the values γ 0 (0), γ 0 (1) equal to γ 0 (0) = lim h↓0

γ(h) − γ(0) , h

γ 0 (1) = lim h↓0

γ(1) − γ(1 − h) . h

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2.5.19 Lemma (Homotopy Lifting Lemma). Under the hypotheses of Theorem 2.5.17, let H(t, s) be a continuous map of [0, 1] × [0, 1] into F such that for each fixed s ∈ [0, 1] the path t 7→ H(t, s) is C 1 . In addition, assume that H(0, s) = y0 , for all s ∈ [0, 1]. If y0 = f (x0 ) for some x0 ∈ E, there exists a unique lift K of H through f which is C 1 in t for every s and K(0, s) = x0 for every s ∈ [0, 1]. See Figure 2.5.5. Proof. Uniqueness follows by Lemma 2.5.18. By the inverse function theorem, there are open neighborhoods U of x0 and V of y0 such that f |U : U → V is a diffeomorphism. Since the open set H −1 (U ) contains the closed set {0} × [0, 1], there exists ε > 0 such that [0, ε[ × [0, 1] ⊂ H −1 (U ). Let K : [0, ε[ × [0, 1] → E be given by K = f −1 ◦ H. Consider the set A = { δ ∈ [0, 1] | H : [0, δ[ × [0, 1] → F can be lifted through f to E and the lift K satisfies K(0, s) = x0 and K is C 1 in t for every s ∈ [0, 1] }, which contains the interval [0, ε[. If α = sup A we shall show first that α ∈ A and second that α = 1. This will prove the existence of the lifting K.

E  * K  

f ? - F

[0, 1] × [0, 1] H

Figure 2.5.5. The homotopy lifting lemma

To show that α ∈ A, note that for 0 ≤ t < α we have f ◦ K = H and thus Df (K(1, s)) ·

∂H ∂K (1, s) = (1, s). ∂t ∂t

Since Df (K(1, s)) is invertible and f is C 1 , the function s ∈ [0, 1] 7→ kDf (K(1, s))−1 k ∈]0, ∞[ is continuous. Let M = sup kDf (K(1, s))−1 k > 0. s∈[0,1]

Hence we get



∂K



≤ M sup ∂H = N.

∂t

∂t t,s∈[0,1] Thus by the mean value inequality, if {tn } is an increasing sequence in A converging to a, kK(tn , s) − K(tm , s)k ≤ N |tn − tm |, for every s ∈ [0, 1], which shows that {K(tn , s)} is a Cauchy sequence in E, uniformly in s ∈ [0, 1]. Let K(α, s) = lim K(tn , s). tn ↑α

By continuity of f and H we have f (K(α, s)) = lim f (K(tn , s)) = lim H(tn , s) = H(α, s), tn ↑α

tn ↑α

which proves that α ∈ A. Next we show that α = 1. If α < 1 consider the curves s 7→ K(α, s) and s 7→ H(α, s) = f (K(α, s)). For each s ∈ [0, 1] choose open neighborhoods Us of K(α, s) and Vs of H(α, s) such that f |Us : Us → Vs is a diffeomorphism. By compactness of the path K(α, s) in s, that is, of the set { K(α, s) | s ∈ [0, 1] }, finitely

2.5 The Inverse and Implicit Function Theorems

119

many of the Us , say U1 , . . . , Un , cover it. Therefore the corresponding V1 , . . . , Vn cover { H(α, s) | s ∈ [0, 1] }. Since H −1 (Vi ) contains the point (α, si ), there exists ε > 0 such that ]α − εi , α + εi [ × ]si − δi , si + δi [ ⊂ H −1 (Vi ), where ]si − δi , si + δi [ ⊂ H(α, ·)−1 (Vi ) and in particular ]si − δi , si + δi [, i = 1, . . . , n cover [0, 1]. Let ˜ : [0, α + ε[ × [0, 1] → E by ε = min{ε1 , . . . , εn } and define K ( K(t, s), if (t, s) ∈ [0, α[ × {(0, 1)}; ˜ K(t, s) = −1 (f |Ui ) (H(t, s)), if (t, s) ∈ [α, α + ε[ × ]si − δi , si + δi [, ˜ is a lifting of H, contradicting the definition of α. where i = 1, . . . , n. Thus, K Since f (K(0, s)) = H(0, s) = y0 for all s ∈ [0, 1], the local injectivity of f , the continuity of s 7→ K(0, s), the compactness of [0, 1] and K(0, 0) = x0 imply that K(0, s) = x0 . Finally, K is C 1 in t for each s by the chain rule: ∂H ∂K = Df (K(t, s))−1 · . ∂t ∂t



2.5.20 Corollary (Path Lifting Lemma). Under the hypotheses of Theorem 2.5.17, let γ : [0, 1] → F be a C 1 path. If γ(0) = y0 = f (x0 ) for some x0 ∈ E, there is a unique C 1 path δ : [0, 1] → E such that f ◦ δ = γ and δ(0) = x0 . Proof. If H(t, s) = γ(t) for all (t, s) ∈ [0, 1]×[0, 1], then by the Homotopy Lifting Lemma there is a unique lift K of H through f . Thus, for each fixed t ∈ [0, 1], f (K(t, s)) = H(t, s) = γ(t) for all s ∈ [0, 1]. Local injectivity of f , the continuity of s 7→ K(t, s) for t fixed, and the compactness of [0, 1], imply that K(t, s) is constant in s. Since this argument is valid for each fixed t ∈ [0, 1], it follows that K(t, s) does not depend on s, that is, K(t, s) = δ(t) is a path satisfying f ◦ δ = γ and δ(0) = K(0, s) = x0 .  Proof of Theorem 2.5.17. Let y0 , y ∈ F and consider the path γ(t) = (1 − t)y0 + ty. The Path Lifting Lemma guarantees the existence of a C 1 path δ : [0, 1] → E lifting γ, that is, f ◦ δ = γ. In particular, f (δ(1)) = γ(1) = y and thus f is surjective. Next we prove that f is injective. Let x1 , x2 ∈ E be such that f (x1 ) = f (x2 ) = y. Define H : [0, 1]×[0, 1] → F by H(t, s) = (1 − s)f (tx1 ) + sf (tx2 ). This map is C 1 in both variables and H(0, s) = f (0) for all s ∈ [0, 1]. By the Homotopy Lifting Lemma, there is a unique continuous lift K : [0, 1] × [0, 1] → E such that f ◦ K = H and K(0, s) = 0 for all s ∈ [0, 1]. Since f is a local C 1 diffeomorphism and H is a C 1 map, it follows that K is a C 1 map in both variables. The s derivative of the identity H(1, s) = (1 − s)f (x1 ) + sf (x2 ) = y yields 0=

d d d H(1, s) = f (K(1, s)) = Df (K(1, s)) · K(1, s). ds ds ds

Since Df (K(1, s)) is an isomorphism, this implies that dK(1, s)/ds = 0 and thus, by integration, K(1, 0) = K(1, 1). Thus, since f is a local diffeomorphism, we have x1 = f −1 (f (x1 )) = f −1 (H(1, 0)) = K(1, 0) = K(1, 1) = f −1 (H(1, 1)) = f −1 (f (x2 )) = x2 , which proves that f is injective. Therefore f is a bijective map which is a local diffeomorphism around every point, that is, f is a diffeomorphism of E with F.  Remarks. (i) The uniform bound on kDf (x)−1 k can be replaced by properness of the map (see Exercise 1.5-10). Indeed, the only place where the uniform bound on kDf (x)−1 k was used is in the homotopy

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lifting lemma in the argument that α = sup A ∈ A. If f is proper, this is shown in the following way. Let {t(n)} be an increasing sequence in A converging to α. Then H(t(n), s) → H(α, s) and from f ◦ K = H on [0, α[ × [0, 1], it follows that f (K(t(n), s)) → H(α, s) uniformly in s ∈ [0, 1]. Thus, by properness of f , there is a subsequence {t(m)} such that K(t(m), s) is convergent for every s. Put K(α, s) = limt(m)↑α K(t(n), s) and proceed as before. (ii) Conditions on f such as properness, or the conditions in the theorem are necessary as the following counterexample shows. Let f : R2 → R2 be given by (ex , ye−x ) so that f (R2 ) is the right open half plane and in particular f is not onto. However  x  e 0 Df (x, y) = −ye−x e−x is clearly an isomorphism for every (x, y) ∈ R2 . But f is neither proper nor does the norm kDf (x, y)−1 k have a uniform bound on R2 . For example, the inverse image of the compact set [0, 1]×{0} is ]−∞, 0]× {0} and kDf (x, y)−1 k = C[e−2x + e2x + y 2 e−2x ]1/2 , which is unbounded x → +∞. (iii) See Wu and Desoer [1972] and Ichiraku [1985] for useful references to the theorem and applications. 

Lax–Milgram Theorem. If E = F = H is a Hilbert space, then the Hadamard–Levy theorem has an important consequence. We have seen that in the case of f : R → R with a uniform bound on 1/|f 0 (x)|, the strong monotonicity of f played a key role in the proof that f is a diffeomorphism. 2.5.21 Definition. such that

Let H be a Hilbert space. A map f : H → H is strongly monotone if there exists a > 0 hf (x) − f (y), x − yi ≥ akx − yk2 .

As in calculus, for differentiable maps strong monotonicity takes on a familiar form. 2.5.22 Lemma. Let f : H → H be a differentiable map of the Hilbert space H onto itself. Then f is strongly monotone if and only if hDf (x) · u, ui ≥ akuk2 for some a > 0. Proof. If f is strongly monotone, hf (x + tu) − f (x), tui ≥ at2 kuk2 for any x, u ∈ H, t ∈ R. Dividing by t2 and taking the limit as t → 0 yields the result. Conversely, integrating both sides of hDf (x + tu) · u, ui ≥ akuk2 from 0 to 1 gives the strong monotonicity condition.  2.5.23 Theorem (Lax–Milgram Theorem). Let H be a real Hilbert space and A ∈ L(H, H) satisfy the estimate hAe, ei ≥ akek2 for all e ∈ H. Then A is an isomorphism and kA−1 k ≤ 1/a. Proof. The condition clearly implies injectivity of A. To prove A is surjective, we show first that A(H) is closed and then that the orthogonal complement A(H)⊥ is {0}. Let fn = A(en ) be a sequence which converges to f ∈ H. Since kAek ≥ akek by the Schwarz inequality, we have kfn − fm k = kA(en − em )k ≥ aken − em k, and thus {en } is a Cauchy sequence in H. If e is its limit we have Ae = f and thus f ∈ A(H). To prove A(H)⊥ = {0}, let u ∈ A(H)⊥ so that 0 = hAu, ui ≥ akuk2 whence u = 0. By Banach’s isomorphism theorem 2.2.16, A is a Banach space isomorphism of H with itself. Finally, replacing e by A−1 f in kAek ≥ akek yields kA−1 f k ≤ kf k/a, that is, kA−1 k ≤ 1/a. 

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121

Lemma 2.5.22, the Lax-Milgram Theorem, and the Hadamard–Levy Theorem imply the following global inverse function theorem on a real Hilbert space. 2.5.24 Theorem. Let H be a real Hilbert space and f : H → H be a strongly monotone C k mapping k ≥ 1. Then f is a C k diffeomorphism.

Supplement 2.5E Higher-Order Derivatives of the Inversion Map Let E and F be isomorphic Banach spaces and consider the inversion map I : GL(E, F) → GL(F, E); I(ϕ) = ϕ−1 . We have shown that I is C ∞ and DI(ϕ) · ψ = −ϕ−1 ◦ ψ ◦ ϕ−1 for ϕ ∈ GL(E, F) and ψ ∈ L(E, F). We shall give below the formula for Dk I. The proof is straightforward and done by a simple induction argument that will be left to the reader. Define the map αk+1 : L(F, E) × · · · × L(F, E) {there are k + 1 factors} → Lk (L(E, F); L(F, E)) by αk+1 (χ1 , . . . , χk+1 ) · (ψ1 , . . . , ψk ) = (−1)k χ1 ◦ ψ1 ◦ χ2 ◦ ψ2 ◦ · · · ◦ χk ◦ ψk ◦ χk+1 , where χi ∈ L(F, E), i = 1, . . . , k + 1 and ψj ∈ L(E, F), j = 1, ..., k. Let I × · · · × I {with k + 1 factors} be the mapping of GL(E, F) to GL(F, E) × · · · × GL(F, E) with {k + 1 factors} defined by (I × · · · × I)(ϕ) = (ϕ−1 , . . . , ϕ−1 ). Then Dk I = k! Symk ◦ αk+1 ◦ (I × · · · × I), where Symk denotes the symmetrization operator. Explicitly, for ϕ ∈ GL(E, E)

and ψ1 , . . . , ψk ∈ L(E, F),

this formula becomes Dk I(ϕ) · (ψ1 , . . . , ψk ) = (−1)k

X

ϕ−1 ◦ ψσ(1) ◦ ϕ−1 · · · ◦ ϕ−1 ◦ ψσ(k) ◦ ϕ−1 ,

σ∈Sk

where Sk is the group of permutations of {1, . . . , k} (see Supplements 2.2B and 2.4A).

Exercises  2.5-1.

Let f : R4 → R2 be defined by f (x, y, u, v) = (u3 + vx + y, uy + v 3 − x).

At what points can we solve f (x, y, u, v) = (0, 0) for (u, v) in terms of (x, y)? Compute ∂u/∂x.

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2. Banach Spaces and Differential Calculus

(i) Let E be a Banach space. Using the inverse function theorem, show that each A in a neigh 2.5-2. borhood of the identity map in GL(E, E) has a unique square root. (ii) Show that for A ∈ L(E, E) the series 1 1 B = 1 − (I − A) − 2 (I − A)2 − · · · 2 2 2! 1 · 3 · 5 · · · (2n − 3) − (I − A)n − · · · 2n n! is absolutely convergent for kI − Ak < 1. Check directly that B 2 = A.  2.5-3.

(i) Let A ∈ L(E, E) and let eA =

∞ X An . n! n=0

Show this series is absolutely convergent and find an estimate for keA k, A ∈ L(E, E). (ii) Show that if AB = BA, then eA+B = eA eB = eB eA . Conclude that (eA )−1 = e−A ; that is, eA ∈ GL(E, E). (iii) Show that e(·) : L(E, E) → GL(E, E) is analytic (you may regard the exponential map as taking values in the linear space L(E, E); recall that GL(E, E) ⊂ L(E, E) is an open subset. (iv) Use the inverse function theorem to conclude that A 7→ eA has a unique inverse around the origin. Call this inverse A 7→ log A and note that log I = 0. (v) Show that if kI − Ak < 1, the function log A is given by the absolutely convergent power series log A =

∞ X (−1)n−1 (A − I)n . n n=1

(vi) If kI − Ak < 1, kI − Bk < 1, and AB = BA, conclude that log (AB) = log A + log B. In particular, log A−1 = − log A.  2.5-4. Show that the implicit function theorem implies the inverse function theorem. Hint: Apply the implicit function theorem to g : U × F → F, g(u, v) = f (u) − v, for f : U ⊂ E → F.  2.5-5.

Let f : R2 → R2 be C ∞ and satisfy the Cauchy–Riemann equations (see Exercise 2.3-6): ∂f1 ∂f2 = , ∂x ∂y

∂f1 ∂f2 =− . ∂y ∂x

Show that Df (x, y) = 0 iff det(Df (x, y)) = 0. Show that the local inverse (where it exists) also satisfies the Cauchy–Riemann equations. Give a counterexample for the first statement, if f does not satisfy Cauchy– Riemann.  2.5-6.

Let f : R → R be given by f (x) = x + x2 cos

Show that (i)

f is continuous;

(ii) f is differentiable at all points;

1 x

if x 6= 0,

and f (0) = 0.

2.5 The Inverse and Implicit Function Theorems

123

(iii) the derivative is discontinuous at x = 0; (iv) f 0 (0) 6= 0; (v) f has no inverse in any neighborhood of x = 0. (This shows that in the inverse function theorem the continuity hypothesis on the derivative cannot be dropped.)  2.5-7. It is essential to have Banach spaces in the inverse function theorem rather than more general spaces such as topological vector spaces or Fr´echet spaces. (The following example of the failure of Theorem 2.5.2 in Fr´echet spaces is due to M. McCracken.) Let H(∆) denote the set of all analytic functions on the open unit disk in C, with the topology of uniform convergence on compact subsets. Let F : H(∆) → H(∆) be defined by ∞ X

an z n 7→

n=0

Show that F is C

∞

∞ X

a2n z n .

n=0

and that DF

∞ X

! an z

n

·

n=0

∞ X

! bn z

n

n=1

=

∞ X

2 an bn z n .

n=1

(Define the Fr´echet derivative in H(∆) as part of your answer.) If a0 = 1 and an = 1/n, n 6= 1, then ! ∞ X zn DF n n=1 is a bounded linear isomorphism. However, since   z k−1 z2 zk z k+1 F z+ + ··· + − + + ··· = F 2 k−1 k k+1

∞ X zn n n=1

!

conclude that F is not locally injective. (Consult Schwartz [1967], Sternberg [1969], and Hamilton [1982] for more sophisticated versions of the inverse function theorem valid in Fr´echet spaces.)  2.5-8 (Generalized Lagrange Multiplier Theorem; Luenberger [1969]). Let f : U ⊂ E → F and g : U ⊂ E → G be C 1 and suppose Dg(u0 ) is surjective. Suppose f has a local extremum (maximum or minimum) at u0 subject to the constraint g(u) = 0. Then prove (i) Df (u0 ) · h = 0 for all h ∈ ker Dg(u0 ), and (ii) there is a λ ∈ G∗ such that Df (u0 ) = λDg(u0 ). (See Supplement 3.5A for the geometry behind this result).  2.5-9.

Let f : U ⊂ Rm → Rn be a C 1 map.

(i) Show that the set Gr = { x ∈ U | rank Df (x) ≥ r } is open in U . Hint: If x0 ∈ Gr , let M (x0 ) be a square block of the matrix of Df (x0 ) in given bases of Rm and Rn of size ≥ r such that det M (x0 ) 6= 0. Using continuity of the determinant function, what can you say about det M (x) for x near x0 ? (ii) We say that R is the maximal rank of Df (x) on U if R = sup (rank Df (x)). x∈U

Show that VR = { x ∈ U | rank Df (x) = R } is open in U . Conclude that if rank Df (x0 ) is maximal then rank Df (x) stays maximal in a neighborhood of x0 .

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2. Banach Spaces and Differential Calculus

(iii) Define Oi = int{ x ∈ U | rank Df (x) = i } and let R be the maximal rank of Df (x), x ∈ U . Show that O0 ∪ · · · ∪ OR is dense in U . Hint: Let x ∈ U and let V be an arbitrary neighborhood of x. If Q denotes the maximal rank of Df (x) on x ∈ V , use (ii) to argue that V ∩ OQ = { x ∈ V | rank Df (x) = Q } is open and nonempty in V . (iv) Show that if a C 1 map f : U ⊂ Rm → Rn is injective (surjective onto an open set), then m ≤ n (m ≥ n). Hint: Use the rank theorem and (ii).  2.5-10 (Uniform Contraction Principle; Hale [1969], Chow and Hale [1982]). (i) Let T : cl(U ) × V → E be a C k map, where U ⊂ E and V ⊂ F are open sets. Suppose that for fixed y ∈ V , T (x, y) is a contraction in x, uniformly in y. If g(y) denotes the unique fixed point of T (x, y), show that g is C k . Hint: Proceed directly as in the proof of the inverse mapping theorem. (ii) Use (i) to prove the inverse mapping theorem.  2.5-11 (Lipschitz Inverse Function Theorem; Hirsch and Pugh [1970]). (i) Let (Xi , di ) be metric spaces and f : X1 → X2 . The map f is called Lipschitz if there exists a constant L such that d2 (f (x), f (y)) ≤ Ld1 (x, y) for all x, y ∈ X1 . The smallest such L is the Lipschitz constant L(f ). Thus, if X1 = X2 and L(f ) < 1, then f is a contraction. If f is not Lipschitz, set L(f ) = ∞. Show that if g : (X2 , d2 ) → (X3 , d3 ), then L(g ◦ f ) ≤ L(g)L(f ). Show that if X1 , X2 are normed vector spaces and f, g : X1 → X2 , then L(f + g) ≤ L(f ) + L(g),

L(f ) − L(g) ≤ L(f − g).

(ii) Let E be a Banach space, U an open set in E such that the closed ball Br (0) ⊂ U . Let f : U → E be given by f (x) = x + ϕ(x), where ϕ(0) = 0 and ϕ is a contraction. Show that f (Dr (0)) ⊃ Dr(1−L(ϕ)) (0), that f is invertible on f −1 (Dr(1−L(ϕ)) (0)), and that f −1 is Lipschitz with constant L(f −1 ) ≤ 1/(1 − L(ϕ)). Hint: If kyk < r(1 − L(ϕ)), define F : U → E by F (x) = y − ϕ(x). Apply the contraction mapping principle in Br (0) and show that the fixed point is in Dr (0). Finally, note that (1 − L(ϕ))kx1 − x2 k ≤ kx1 − x2 k − kϕ(x1 ) − ϕ(x2 )k ≤ kf (x1 ) − f (x2 )k. (iii) Let U be an open set in the Banach space E, V be an open set in the Banach space F, x0 ∈ U , Br (x0 ) ⊂ U . Let α : U → V be a homeomorphism. Assume that α−1 : V → U is Lipschitz and let ψ : U → F be another Lipschitz map. Assume L(ψ)L(α−1 ) < 1 and define f = α + ψ : U → F. Denote y0 = f (x0 ). Show that f (α−1 (Dr (x0 ))) ⊃ Dr(1−L(ψ)L(α−1 )) (y0 ), that f is invertible on f −1 (Dr(1−L(ψ)L(α−1 )) (y0 ), and that f −1 is Lipschitz with constant L(f −1 ) ≤

1 L(α−1 )−1

− L(ψ)

.

Hint: Replacing ψ by the map x 7→ ψ(x)−ψ(x0 ) and V by V +{ψ(x0 )}, we can assume that ψ(x0 ) = 0 and f (x0 ) = α(x0 ) = y0 . Next, replace this new f by x 7→ f (x + x0 ) − f (x0 ), U by U − {x0 }, and the new V by V + {y0 }; thus we can assume that x0 = 0,

y0 = 0,

ψ(0) = 0,

and α(0) = 0.

2.5 The Inverse and Implicit Function Theorems

125

Then f ◦ α−1 = I + ψ ◦ α−1 , (ψ ◦ α−1 )(0) = 0, L(ψ ◦ α−1 ) ≤ L(ψ)L(α−1 ) < 1, so (ii) is applicable. (iv) Show that |L(f −1 ) − L(α−1 )| → 0 as L(ψ) → 0. Let α : R → R be the homeomorphism defined by α(x) = x if x ≤ 0 and α(x) = 2x if x ≥ 0. Show that both α and α−1 are Lipschitz. Let ψ(x) = c = constant. Show that L(ψ) = 0 and if c 6= 0, then L(f −1 − α−1 ) ≥ 1/2. Prove, however, that if α, f are diffeomorphisms, then L(f −1 − α−1 ) → 0 as L(ψ) → 0.  2.5-12. Use the implicit function theorem to show that simple roots of polynomials are smooth functions of their coefficients. Conclude that simple eigenvalues of operators of Rn are smooth functions of the operator. Hint: If p(t) = an tn + an−1 tn−1 + · · · + a0 , define a smooth map F : Rn+2 → R by F (an , . . . , a0 , λ) = p(λ) and note that if λ0 is a simple eigenvalue, ∂F (λ0 )/∂λ 6= 0.  2.5-13. Let E, F be Banach spaces, f : U → V a C r bijective map, r ≥ 1, between two open sets U ⊂ E, V ⊂ F. Assume that for each x ∈ U , Df (x) has closed split image and is one-to-one. (i) Use the local immersion theorem to show that f is a C r diffeomorphism. (ii) What fails for the function y = x3 ?  2.5-14. Let E be a Banach space, U ⊂ E open and f : U → R a C r map, r ≥ 2. We say that u ∈ U is a critical point of f , if Df (u) = 0. The critical point u is called strongly non-degenerate if D2 f (u) induces a Banach space isomorphism of E with its dual E∗ . Use the Inverse Function Theorem on Df to show that strongly non-degenerate points are isolated, that is, each strongly non-degenerate point is unique in one of its neighborhoods. (A counter-example, if D2 f is only injective, is given in Exercise 2.4-15.)  2.5-15.

For u : S 1 → R, consider the equation du 1 + u2 − dθ 2π

Z

2π

u2 dθ = ε sin θ

0

where θ is a 2π-periodic angular variable and ε is a constant. Show that if ε is sufficiently small, this equation has a solution.  2.5-16.

Use the implicit function theorem to study solvability of ∇2 ϕ + ϕ3 = f in Ω

and

∂ϕ = g on ∂Ω, ∂n

where Ω is a region in Rn with smooth boundary, as in Supplement 2.5C.  2.5-17.

Let E be a finite dimensional vector space.

(i) Show that det(exp A) = etraceA . Hint: Show it for A diagonalizable and then use Exercise 2.2-12(i). (ii) If E is real, show that exp(L(E, E)) ∩ { A ∈ GL(E) | det A < 0 } = ∅. This shows that the exponential map is not onto.

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(iii) If E is complex, show that the exponential map is onto. For this you will need to recall the following facts from linear algebra. Let p be the characteristic polynomial of A ∈ L(E, E), that is, p(λ) = det(A − λI). Assume that p has m distinct roots λ1 , . . . , λm such that the multiplicity of λi is ki . Then E=

m M

ker(A − λi I)ki

and

dim(ker(A − λi I)ki ) = ki

i=1

Thus, to prove the exponential is onto, it suffices to prove it for operators S ∈ GL(E) for which the characteristic polynomial is (λ − λ0 )k . Hint: Since S is invertible, λ0 6= 0, so write λ0 = ez , z ∈ C. Let N = λ−1 0 S − I and let M=

k−1 X i=1

(−1)i−1 N i . i

By the Cayley–Hamilton theorem (see Exercise 2.2-12(ii)), N k = 0, and from the fact that exp(log(1 + w)) = 1 + w for all w ∈ C, it follows that exp(M + zI) = λ0 exp M = λ0 (I + N ) = S.

Page 127

3 Manifolds and Vector Bundles

This version: January 18, 2007

We are now ready to study manifolds and the differential calculus of maps between manifolds. Manifolds are an abstraction of the idea of a smooth surface in Euclidean space. This abstraction has proved useful because many sets that are smooth in some sense are not presented to us as subsets of Euclidean space. The abstraction strips away the containing space and makes constructions intrinsic to the manifold itself. This point of view is well worth the geometric insight it provides.

3.1

Manifolds

Historical Remarks. The history of manifolds is a complex one, so we confine ourselves to a few brief remarks to help orient the reader. The story starts with the development of noneuclidean geometry when it was realized that there were spaces other than standard Euclidean spaces that are, at least mathematically, interesting. This took place in the early 1800’s and was primarily due to Janos Bolyai (1802–1860), Carl Friedrich Gauss (1777-1855) and Nikolai Invanovich Lobachevsky (1792–1856). Around this time, surfaces were generally regarded simply as sets described by sets of algebraic equations. The next main event was the 1854 Inagaural Lecture of George Bernhard Riemann, On the hypotheses underlying the foundations of geometry, in which he set out to generalize Gauss’ theory of surfaces and which was the main inspiration for subsequent developments. He imagined a space of many dimensions carrying the infinitesimal notion of length. His examples included the elliptic and hyperbolic spaces of Bolyai, Gauss and Lobachevsky, along with many others. Albert Einstein along with Marcel Grossman were motivated very much by Riemann’s ideas in the creation of general relativity in the early 1900’s. But the period between Riemann and the creation of general relativity, had much more activity as well. First of all, building on ideas of Felix Klein, who laid the foundations for connecting group theory and noneuclidean geometries (the so-called Erlangen Program of Klein), Sophus Lie1 wrote his monumental three volume treatise on Lie groups, which appeared in the era around 1890. As we shall see in Chapter 5, Lie groups are intimately connected with the theory of manifolds. 1 Pronounced

“Lee”.

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Another major player in the development was Henri Poincar´e (1854–1912) who created the subject of algebraic topology as a tool to help him in his studies of dynamics and mechanics, such as the 3-body problem. Poincar´e of course needed spaces on which to hang his algebraic topology; for him these were things such as level sets of functions, a class of examples that is still key (we shall study this class of manifolds in §3.5). While Poincar´e was mainly dealing with concrete examples, there remained the problem of providing an abstract definition of a manifold. For example, should they somehow be defined in terms of subsets of Rn or be given more abstractly? It was not clear at the time that one could give a notion of a manifold without it being “inside” some containing space. The modern definition indeed dispenses with the containing space, although a theorem of Whitney finally tells one that all abstract manifolds can indeed be realized as subsets of high dimensional Euclidean spaces. The modern abstract definition was provided by the axiomitization ideas of David Hilbert (1862-1943) that were then improved by H. Weyl (1885-1955) in his 1913 work Der Idee der Riemannschen Fl¨ ache that more or less gives the abstract definition that we study in the next section. Elie Joseph Cartan and Charles Ehresmann then did deep work on merging the ideas of manifolds and Lie groups. Of course there have been many modern developments, but we shall stop at this point. Charts and Atlases. The basic idea of a manifold is to introduce a local object that will support differentiation processes and then to patch these local objects together smoothly. Before giving the formal definitions it is good to have an example in mind. In Rn+1 consider the n-sphere S n ; that is, the set of x ∈ Rn+1 such that kxk = 1 (k · k denotes the usual Euclidean norm). We can construct bijections from subsets of S n to Rn in several ways. One way is to project stereographically from the south pole onto the hyperplane that cuts through the equator of the sphere. This is a bijection from S n , with the south pole removed, onto Rn . Similarly, we can interchange the roles of the poles to obtain another bijection. (See Figure 3.1.1.)

N

P

S Figure 3.1.1. The two-sphere S 2 .

With the usual relative topology on S n as a subset of Rn+1 , each of these maps is a homeomorphisms from their domain to Rn . Each map takes the sphere (minus the appropriate pole) to an open subset of Rn . If we go from Rn to the sphere by one map, then back to Rn by the other, we get a smooth map from an open subset of Rn to Rn . Each map assigns a coordinate system to S n minus a pole. The union of the two domains is S n , but no single homeomorphism can be used between S n and Rn ; however, we can cover S n using two of them. In this case they are compatible; that is, in the region covered by both coordinate systems, the change of coordinates is smooth. For some studies of the sphere, and for other manifolds, two coordinate systems will not suffice. We thus allow all other coordinate systems compatible with these. For example, on S 2 we want to allow spherical coordinates (θ, ϕ) since they are convenient for many computations. With this example in mind, we are ready to state the general definition.

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129

3.1.1 Definition. Let S be a set. A chart on S is a bijection ϕ from a subset U of S to an open subset of a Banach space. We sometimes denote ϕ by (U, ϕ), to indicate the domain U of ϕ. A C k atlas on S is a family of charts A = { (Ui , ϕi ) | i ∈ I } such that MA1. S =

S

{ Ui | i ∈ I }.

MA2. Any two charts in A are compatible in the sense that the overlap maps between members of A are C k diffeomorphisms: for two charts (Ui , ϕi ) and (Uj , ϕj ) with Ui ∩ Uj 6= ∅, we form the −1 −1 overlap map: ϕji = ϕj ◦ϕ−1 i |ϕi (Ui ∩Uj ), where ϕi |ϕi (Ui ∩Uj ) means the restriction of ϕi k to the set ϕi (Ui ∩Uj ). We require that ϕi (Ui ∩Uj ) is open and that ϕji be a C diffeomorphism. (See Figure 3.1.2.)

Figure 3.1.2. Charts ϕi and ϕj on a manifold

3.1.2 Examples. A.

Any Banach space F admits an atlas formed by the single chart (F, identity).

B. A less trivial example is the atlas formed by the two charts of S n discussed previously. More explicitly, if N = (0, 0, . . . , 0, 1) and S = (0, 0, . . . , 0, −1) are the north and south poles of S n , the stereographic projections from N and S are n

n

ϕ1 : S \{N } → R ,



1

n+1

1

n+1

ϕ1 (x , . . . , x

)=

xn x1 . . . , , 1 − xn+1 1 − xn+1



x1 xn , . . . , 1 + xn+1 1 + xn+1



,

and n

n

ϕ2 : S \{S} → R ,

ϕ2 (x , . . . , x

 )=

2 and the overlap map ϕ2 ◦ ϕ−1 : Rn \{0} → Rn \{0} is given by the mapping (ϕ2 ◦ ϕ−1 1 1 )(z) = z/kzk , n ∞ n z ∈ R \{0}, which is clearly a C diffeomorphism of R \{0} to itself. 

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3. Manifolds and Vector Bundles

Definition of a Manifold.

We are now ready for the formal definition of a manifold.

k

3.1.3 Definition. Two C atlases A1 and A2 are equivalent if A1 ∪A2 is a C k atlas. A C k differentiable structure D on S is an equivalence class of atlases on S. The union of the atlases in D, [ AD = { A | A ∈ D }, is the maximal atlas of D, and a chart (U, ϕ) ∈ AD is an admissible local chart. If A is a C k atlas on S, the union of all atlases equivalent to A is called the C k differentiable structure generated by A. A differentiable manifold M is a pair (S, D), where S is a set and D is a C k differentiable structure on S. We shall often identify M with the underlying set S for notational convenience. If a covering by charts takes their values in a Banach space E, then E is called the model space and we say that M is a C k Banach manifold modeled on E. If no differentiability class is explicitly given, a manifold will be assumed to be C ∞ (also referred to as “smooth”). If we make a choice of a C k atlas A on S then we obtain a maximal atlas by including all charts whose overlap maps with those in A are C k . In practice it is sufficient to specify a particular atlas on S to determine a manifold structure for S. 3.1.4 Example. An alternative atlas for S n has the following 2(n + 1) charts: (Ui± , ψi± ), i = 1, . . . , n + 1, where Ui± = { x ∈ S n | ±xi > 0 } and ψi± : Ui± → { y ∈ Rn | kyk < 1 } is defined by ψi± (x1 , . . . , xn+1 ) = (x1 , . . . , xi−1 , xi+1 , . . . , xn+1 ); ψi± projects the hemisphere containing the pole (0, . . . , ±1, . . . , 0) onto the open unit ball in the tangent space to the sphere at that pole. It is verified that this atlas and the one in Example 3.1.2B with two charts are equivalent. The overlap maps of this atlas are given by 

−1  1 y , . . . , yn ψj± ◦ ψi±   p = y 1 , . . . , y j−1 , y j+1 , . . . , y i−1 , ± 1 − kyk2 , y i , . . . , y n , where j < i.



Topology of a Manifold.

We now define the open subsets in a manifold, which will give us a topology.

3.1.5 Definition. Let M be a differentiable manifold. A subset A ⊂ M is called open if for each a ∈ A there is an admissible local chart (U, ϕ) such that a ∈ U and U ⊂ A. 3.1.6 Proposition. Proof.

The open sets in M define a topology.

Take as basis of the topology the family of finite intersections of chart domains.



3.1.7 Definition. A differentiable manifold M is an n-manifold when every chart has values in an ndimensional vector space. Thus for every point a ∈ M there is an admissible local chart (U, ϕ) with a ∈ U and ϕ(U ) ⊂ Rn . We write n = dim M . An n-manifold will mean a Hausdorff, differentiable n-manifold in this book. A differentiable manifold is called a finite-dimensional manifold if its connected components are all n-manifolds (n can vary with the component). A differentiable manifold is called a Banach manifold (resp. Hilbert manifold ) if the model space is a Banach (resp. Hilbert) space.2 No assumption on the connectedness of a manifold has been made. In fact, in some applications the manifolds are disconnected (see Exercise 3.1-3). Since manifolds are locally arcwise connected, their components are both open and closed. 2 One

can similarly form a manifold modeled on any linear space in which one has a theory of differential calculus. For example mathematicians often speak of a “Fr´ echet manifold,” a “LCTVS manifold,” etc. We have chosen to stick with Banach manifolds here primarily to avail ourselves of the inverse function theorem. See Exercise 2.5-7.

3.1 Manifolds

131

3.1.8 Examples. A. Every discrete topological space S is a 0-manifold, the charts being given by the pairs ({s}, ϕs ), where ϕs : s 7→ 0 and s ∈ S. B. Every Banach space is a manifold; its differentiable structure is given by the atlas with the single identity chart. C. The n-sphere S n with a maximal atlas generated by the atlas with two charts described in Examples 3.1.2B or 3.1.4 makes S n into an n-manifold. The reader can verify that the resulting topology is the same as that induced on S n as a subset of Rn+1 . D. A set can have more than one differentiable structure. For example, R has the following incompatible charts: (U1 , ϕ1 ) : U1 = R, (U2 , ϕ2 ) : U2 = R,

ϕ1 (r) = r3 ∈ R; ϕ2 (r) = r ∈ R.

and

They are not compatible since ϕ2 ◦ ϕ−1 1 is not differentiable at the origin. Nevertheless, these two structures are “diffeomorphic” (Exercise 3.2-8), but structures can be “essentially different” on more complicated sets (e.g., S 7 ). That S 7 has two nondiffeomorphic differentiable structures is a famous result of Milnor [1956]. Similar phenomena have been found on R4 by Donaldson [1983]; see also Freed and Uhlenbeck [1984]. E. Essentially the only one-dimensional paracompact connected manifolds are R and S 1 . This means that all others are diffeomorphic to R or S 1 (diffeomorphic will be precisely defined later). For example, the circle with a knot is diffeomorphic to S 1 . (See Figure 3.1.3.) See Milnor [1965] or Guillemin and Pollack [1974] for proofs.

= Figure 3.1.3. The knot and circle are diffeomorphic

F. A general two-dimensional compact connected manifold is the sphere with “handles” (see Figure 3.1.4). This includes, for example, the torus, whose precise definition will be given in the next section. This classification of two-manifolds is described in Massey [1991] and Hirsch [1976]. G. Grassmann Manifolds. Let Gn (Rm ), where m ≥ n denote the space of all n-dimensional subspaces of Rm . For example, G1 (R3 ), also called projective 2-space, is the space of all lines in Euclidean three space. The goal of this example is to show that Gn (Rm ) is a smooth compact manifold. In the following supplement, we shall develop an infinite dimensional version of this example. The idea of the proof is the following.3 Replace Rm by an m-dimensional vector space E and let F ⊂ E be a fixed n-dimensional subspace. Choose a fixed complement G of F, so G is an m − n dimensional subspace. The decomposition E = F ⊕ G defines a projection map πF : E → G that is the identity on G and such that 3 This

proof will be fleshed out in future versions of the book.

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3. Manifolds and Vector Bundles

Figure 3.1.4. The sphere with handles

F is the kernel of πF . Now consider the affine space of all projections of E to G that are the identity on G. Subspaces near F of dimension n are in one to one correspondence with projections that are the identity on G by associating to each such projection, its kernel. But these projections are an affine subspace of L(E, G), so can be identified with an open set in a linear space. This association is in fact a chart for Gn (Rm ). It takes a little work to show that overlap maps are smooth, but this is straightforward. In fact, Gn (Rm ) is a compact and connected n(m − n)-manifold. To prove compactness, consider the set Fn,m of oriented orthogonal sets of n unit vectors in Rm . Since Fn,m is closed and bounded in Rm × · · · × Rm (n times), Fn,m is compact by the Heine-Borel theorem. Thus Gn (Rm ) is compact, since it is the continuous image of Fn,m by the map {e1 , . . . , en } 7→ span {e1 , . . . , en }. Since Fn,m is also connected, so is Gn (Rm ). H. Projective spaces Let RPn = G1 (Rn+1 ) = the set of lines in Rn+1 . Thus from the previous example, RPn is a compact connected real n-manifold. Similarly CPn , the set of complex lines in Cn+1 , is a compact connected (complex) n-manifold. There is a projection π : S n → RPn defined by π(x) = span(x), which is a diffeomorphism restricted to an open hemisphere. Thus, any chart for S n produces one for RPn as well. 

Supplement 3.1A Infinite Dimensional Grassmannians Let E be a Banach space and consider the set G(E) of all split subspaces of E. For F ∈ G(E), let G denote one of its complements, that is, E = F ⊕ G. Let UG = { H ∈ G(E) | E = H ⊕ G }, and, noting that F ∈ UG , define ϕF,G : UG → L(F, G)

by ϕF,G (H) = πF (H, G) ◦ πG (H, F)−1 ,

where πF (G) : E → G, πG (F) : E → F denote the projections induced by the direct sum decomposition E = F ⊕ G, and πF (H, G) = πF (G)|H, πG (H, F) = πG (F)|H. The inverse appearing in the definition of ϕF,G exists as the following argument shows. If H ∈ UG , that is, if E = F ⊕ G = H ⊕ G, then the maps πG (H, F) ∈ L(H, F) and πG (F, H) ∈ L(F, H) are invertible and one is the inverse of the other, for if h = f + g, then f = h − g, for f ∈ F, g ∈ G, and h ∈ H, so

3.1 Manifolds

133

that (πG (F, H) ◦ πG (H, F))(h) = πG (F, H)(f ) = h, and (πG (H, F) ◦ πG (F, H))(f ) = πG (H, F)(h) = f . In particular, ϕF,G has the alternative expression ϕF,G (H) = πF (H, G) ◦ πG (F, H). Note that we have shown that H ∈ UG implies πG (H, F) ∈ L(H, F) = πG (F)|H ∈ L(H, F) is an isomorphism. The converse is also true, that is, if πG (H, F) is an isomorphism for some split subspace H of E then E = H ⊕ G. Indeed, if x ∈ H ∩ G, then πG (H, F)(x) = 0 and so x = 0, that is, H ∩ G = {0}. If e ∈ E, then we can write   h   i −1 −1 e = πG (H, F) ◦ πG (F) (e) + e − πG (H, F) ◦ πG (F) (e) −1

with the first summand an element of H. Since πG (F)◦(πG (H, F)) is the identity on F, we have πG (F)[e− (πG (H, F)−1 ◦ πG (F))(e)] = 0, that is, the second summand is an element of G and thus E = H + G. Therefore E = H ⊕ G and we have the alternative definition of UG as UG = { H ∈ G(E) | πG (H, F) is an isomorphism of H with F }. Let us next show that ϕF,G : UG → L(F, G) is bijective. For α ∈ L(F, G) define the graph of α by ΓF,G (α) = { f + α(f ) | f ∈ F } which is a closed subspace of E = F ⊕ G. Then E = ΓF,G (α) ⊕ G, that is, ΓF,G (α) ∈ UG , since any e ∈ E can be written as e = f + g = (f + α(f )) + (g − α(f )) for f ∈ F and g ∈ G, and also ΓF,G (α) ∩ G = {0} since f + α(f ) ∈ G for f ∈ F iff f ∈ F ∩ G = {0}. We have ϕF,G (ΓF,G (α)) = πF (ΓF,G (α), G) ◦ πG (F, ΓF,G (α)) which equals the linear map f = (f + α(f )) − α(f ) 7→ πF (ΓF,G (α), G)(f + α(f )) = α(f ) that is, ϕF,G ◦ ΓF,G = identity on L(F, G). Similarly, ΓF,G (πF (H, G) ◦ πG (F, H))ΓF,G (π) = { f + (πF (H, G) ◦ πG (F, H))(f ) | f ∈ F } = { f + πF (H, G)(h) | f ∈ F, f = h + g, h ∈ H, and g ∈ G } = { f − g | f ∈ F, f = h + g, h ∈ H, and g ∈ G } = H, that is, ΓF,G ◦ ϕF,G = identity on UG . Thus, ϕF,G is a bijective map which sends H ∈ UG to an element of L(F, G) whose graph in F ⊕ G is H. We have thus shown that (UG , ϕF,G ) is a chart on G(E). To show that { (UG , ϕF,G ) | E = F ⊕ G } is an atlas on G(E), note that [

[

UG = G(E),

F∈G(E) G

where the second union is taken over all G ∈ G(E) such that E = F ⊕ G for each fixed F ∈ G(E). Thus, MA1 is satisfied. To prove MA2, let (UG0 , ϕF0 ,G0 ) be another chart on G(E) with UG ∩ UG0 6= ∅. We ∞ need to show that ϕF,G (UG ∩ UG0 ) is open in L(F, G) and that ϕF,G ◦ ϕ−1 diffeomorphism of F0 ,G0 is a C 0 0 L(F , G ) to L(F, G). Step 1. Proof of the openness of ϕF,G (UG ∩ UG0 ). Let α ∈ ϕF,G (UG ∩ UG0 ) ⊂ L(F, G) and let H = ΓF,G (α). Then E = H ⊕ G = H ⊕ G0 . Assume for the moment that we can show the existence of an ε > 0 such that if β ∈ L(H, G) and kβk < ε, then ΓH,G (β) ⊕ G0 = E. Then if α0 ∈ L(F, G) is such that kα0 k < ε/kπG (H, F)k, we get ΓH,G (α0 ◦ πG (H, F)) ⊕ G0 = E. We shall prove that ΓF,G (α + α0 ) =

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3. Manifolds and Vector Bundles

ΓH,G (α0 ◦ πG (H, F)). Indeed, since the inverse of πG (H, F) ∈ GL(H, F) is I + α, where I is the identity mapping on F, for any h ∈ H, πG (H, F)(h) + ((α + α0 ) ◦ πG (H, F))(h) = [(I + α) ◦ πG (H, F)](h) + (α0 ◦ πG (H, F))(h) = h + (α0 ◦ πG (H, F))(h), whence the desired equality between the graphs of α + α0 in F ⊕ G and α0 ◦ πG (H, F) in H ⊕ G. Thus we have shown that ΓF,G (α + α0 ) ⊕ G0 = E. Since we always have ΓF,G (α + α0 ) ⊕ G = E (since ΓF,G is bijective with range UG ), we conclude that α + α0 ∈ ϕF,G (UG ∩ UG0 ) thereby proving openness of ϕF,G (UG ∩ UG0 ). To complete the proof of Step 1 we therefore have to show that if E = H ⊕ G = H ⊕ G0 then there is an ε > 0 such that for all β ∈ L(H, G) satisfying kβk < ε, we have ΓH,G (β) ⊕ G0 = E. This in turn is a consequence of the following statement: if E = H ⊕ G = H ⊕ G0 then there is an ε > 0 such for all β ∈ L(H, G) satisfying kβk < ε, we have πG0 (ΓH,G (β), H) ∈ GL(ΓH,G (β), H). Indeed, granted this last statement, write e ∈ E as e = h + g 0 , for some h ∈ H and g 0 ∈ G0 , use the bijectivity of πG0 (ΓH,G (β), H) to find an x ∈ ΓH,G (β) such that h = πG0 (ΓH,G (β), H)(x), and note that πG0 (H)(h − x) = 0, that is, h − x = g10 ∈ G0 ; see Figure 3.1.5. Therefore e = x + (g10 + g 0 ) ∈ ΓH,G (β) + G0 . In addition, we also have ΓH,G (β) ∩ G0 = {0}, for if z ∈ ΓH,G (β) ∩ G0 , then πG0 (ΓH,G (β), H)(z) = 0, whence z = 0 by injectivity of the mapping πG0 (ΓH,G (β), H); thus we have shown E = ΓH,G (β) ⊕ G0 .

G

e

x h

H

Figure 3.1.5. Grassmannian charts

Finally, assume that E = H ⊕ G = H ⊕ G0 . Let us prove that there is an ε > 0 such that if β ∈ L(H, G), satisfies kβk < ε, then πG0 (ΓH,G (β), H) ∈ GL(ΓH,G (β), H). Because of the identity πG (H, ΓH,G (β)) = I+β, where I is the identity mapping on H, we have kI − πG0 (H) ◦ πG (H, ΓH,G (β))k = kI − πG0 (H) ◦ (I + β)k = kπG (H0 ) ◦ (I − (I + β))k ≤ kπG0 (H)k kβk < 1 provided that kβk < ε = 1/kπG0 (H)k. Therefore, we get I − (I − πG0 (H) ◦ πG (H, ΓH,G (β))) = πG0 (H) ◦ πG (H, ΓH,G (β)) ∈ GL(H, H). Since πG (H, ΓH,G (β)) ∈ GL(H, ΓH,G (β)) has inverse πG (ΓH,G (β), H), we obtain πG0 (ΓH,G (β), H) = πG0 (H)|ΓH,G (β) = [πG0 (H) ◦ πG (H, (ΓH,G (β))] ◦ πG (ΓH,G (β), H) ∈ GL(ΓH,G (β), H).

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135

Step 2. Proof that the overlap maps are C ∞ . Let (UG , ϕF,G ), (UG0 and ϕF0 ,G0 ) be two charts at the points F, F0 ∈ G(E) such that UG ∩ UG0 6= ∅. If α ∈ ϕF,G (UG ∩ UG0 ), then I + α ∈ GL(F, ΓF,G (α)), where I is the identity mapping on F, and πG0 (ΓF,G (α), F0 ) ∈ GL(ΓF,G (α), F0 ) since ΓF,G (α) ∈ UG ∩ UG0 . Therefore πG0 (F0 ) ◦ (I + α) ∈ GL(F, F0 ) and we get (ϕF0 ,G0 ◦ ϕ−1 F,G )(α) = ϕF0 ,G0 (ΓF,G (α)) = πF0 (ΓF,G (α), G0 ) ◦ πG0 (F0 , ΓF,G (α)) = πF0 (ΓF,G (α), G0 ) ◦ πG0 (F0 , ΓF,G (α)) ◦ πG0 (F0 ) ◦ (I + α) ◦ [πG0 (F0 ) ◦ (I + α)]−1 = πF0 (G0 ) ◦ (I + α) ◦ [πG0 (F0 ) ◦ (I + α)]−1 which is a C ∞ map from ϕF,G (UG ∩ UG0 ) ⊂ L(F, G) to ϕF0 ,G0 (UG ∩ UG0 ) ⊂ L(F0 , G0 ). Since its inverse is β ∈ L(F0 , G0 ) 7→ πF (G) ◦ (I 0 + β) ◦ [πG (F) ◦ (I 0 + β)]−1 ∈ L(F, G), where I 0 is the identity mapping on F0 , it follows that the maps ϕF0 ,G0 ◦ ϕ−1 F,G are diffeomorphisms. Thus, G(E) is a C ∞ Banach manifold, locally modeled on L(F, G). The preceding proof shows that Gn (E), the space of n-dimensional subspaces of E, and Gn (E), the n-codimensional subspaces, are connected components of G(E).

Exercises  3.1-1. Let S = { (x, y) ∈ R2 | xy = 0 }. Construct two “charts” by mapping each axis to the real line by (x, 0) 7→ x and (0, y) 7→ y. What fails in the definition of a manifold?  3.1-2. Let S = ]0, 1[ × ]0, 1[ ⊂ R2 and for each s, 0 < s < 1 let Vs = {s} × ]0, 1[ and let ϕs : Vs → R, (s, t) 7→ t be a candidate chart. Does this collection of candidate charts make S into a one-manifold?  3.1-3. Let S = { (x, y) ∈ R2 | x2 − y 2 = 1 }. Show that the two charts ϕ± : { (x, y) ∈ S | ±x > 0 } → R, ϕ± (x, y) = y define a manifold structure on the disconnected set S.  3.1-4. On the topological space M obtained from [0, 2π] × R by identifying the point (0, x) with (2π, −x), x ∈ R, consider the following two charts: (i) (]0, 2π[ × R, identity), and (ii) (([0, π[ ∪ ]π, 2π[) × R, ϕ), where ϕ is defined by ϕ(θ, x) = (θ, x) if 0 ≤ θ < π and ϕ(θ, x) = (θ − 2π, −x) if π < θ < 2π. Show that these two charts define a manifold structure on M. This manifold is called the M¨ obius band (see Figure 3.4.3 and Example 3.4.10C for an alternative description). Note that the chart (ii) joins 2π to 0 and twists the second factor R, as required by the topological structure of M. (iii) Repeat a construction like (ii) for K, the Klein bottle.  3.1-5 (Compactification of Rn ). Let {∞} be a one point set and let Rnc = Rn ∪ {∞}. Define the charts (U, ϕ) and (U∞ , ϕ∞ ) by U = Rn , ϕ = identity on Rn , U∞ = Rnc \{0}, ϕ∞ (x) = x/kxk2 , if x 6= ∞, and ϕ∞ (x) = 0, if x = ∞. (i) Show that the atlas Ac = {(U, ϕ), (U∞ , ϕ∞ )} defines a smooth manifold structure on Rnc .

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(ii) Show that with the topology induced by Ac , Rnc becomes a compact topological space. It is called the one-point compactification of Rn . (iii) Show that if n = 2, the differentiable structure of R2c = Cc can be alternatively given by the chart (U, ϕ) and the chart (U∞ , ψ∞ ), where ψ∞ (z) = z −1 , if z 6= ∞ and ψ∞ (z) = 0, if z = ∞. (iv) Show that stereographic projection induces a homeomorphism of Rnc with S n .  3.1-6. (i) Define an equivalence relation ∼ on S n by x ∼ y if x = ±y. Show that S n /∼ is homeomorphic with RPn . (ii) Show that (a) eiθ ∈ S 1 7→ e2iθ ∈ S 1 , and (b) (x, y) ∈ S 1 7→ (xy −1 , if y 6= 0 and ∞, if y = 0) ∈ Rc ∼ = S 1 (see Exercise 3.1-5) induce homeomorphisms of S 1 with RP1 . (iii) Show that neither S n nor RPn can be covered by a single chart.  3.1-7. (i) Define an equivalence relation on S 2n+1 ⊂ C2(n+1) by x ∼ y if y = eiθ x for some θ ∈ R. Show 2n+1 S /∼ is homeomorphic to CPn . (ii) Show that (a) (u, v) ∈ S 3 ⊂ C2 7→ (−2u¯ v , |v|2 − |u|2 ) ∈ S 2 , and (b) (u, v) ∈ S 3 ⊂ C2 7→ (uv −1 , if v 6= 0, and ∞, if v = 0) ∈ R2c ∼ = S 2 (see Exercise 3.1-5) induce 2 1 homeomorphisms of S with CP . The map in (a) is called the classical Hopf fibration; it will be studied further in §3.4.  3.1-8 (Flag manifolds). Let F n denote the set of sequences of nested linear subspaces V1 ⊂ V2 ⊂ · · · ⊂ Vn−1 in Rn (or Cn ), where dim Vi = i. Show that F n is a compact manifold and compute its dimension. (Flag manifolds are typified by F n and come up in the study of symplectic geometry and representations of Lie groups.) Hint: Show that F n is in bijective correspondence with the quotient space GL(n)/upper triangular matrices.

3.2

Submanifolds, Products, and Mappings

A submanifold is the nonlinear analogue of a subspace in linear algebra. Likewise, the product of two manifolds, producing a new manifold, is the analogue of a product vector space. The analogue of linear transformations are the C r maps between manifolds, also introduced in this section. We are not yet ready to differentiate these mappings; this will be possible after we introduce the tangent bundle in §3.3. Submanifolds. If M is a manifold and A ⊂ M is an open subset of M , the differentiable structure of M naturally induces one on A. We call A an open submanifold of M . For example, Gn (E), Gn (E) are open submanifolds of G(E) (see Example 3.1.8G). We would also like to say that S n is a submanifold of Rn+1 , although it is a closed subset. To motivate the general definition we notice that there are charts in Rn+1 in which a neighborhood of S n becomes part of the subspace Rn . Figure 3.2.1 illustrates this for n = 1. 3.2.1 Definition. A submanifold of a manifold M is a subset B ⊂ M with the property that for each b ∈ B there is an admissible chart (U, ϕ) in M with b ∈ U which has the submanifold property , namely, that ϕ has the form SM. ϕ : U → E × F,

and

ϕ(U ∩ B) = ϕ(U ) ∩ (E × {0}).

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137

Figure 3.2.1. Submanifold charts for S 1

An open subset V of M is a submanifold in this sense. Here we merely take F = {0}, and for x ∈ V use any chart (U, ϕ) of M for which x ∈ U . 3.2.2 Proposition. Let B be a submanifold of a manifold M . Then B itself is a manifold with differentiable structure generated by the atlas: { (U ∩ B, ϕ|U ∩ B) | (U, ϕ) is an admissible chart in M having property SM for B }. Furthermore, the topology on B is the relative topology. Proof. If Ui ∩ Uj ∩ B 6= ∅, and (Ui , ϕi ) and (Uj , ϕj ) both have the submanifold property, and if we write ϕi = (αi , βi ) and ϕj = (αj , βj ), where αi : Ui → E, αj : Uj → E, βi : Ui → F, and βj : Uj → F, then the maps αi |Ui ∩ B : Ui ∩ B → ϕi (Ui ) ∩ (E × {0}) and αj |Uj ∩ B : Uj ∩ B → ϕj (Uj ) ∩ (E × {0}) are bijective. The overlap map (ϕj |Uj ∩ B) ◦ (ϕi |Ui ∩ B)−1 is given by (e, 0) 7→ ((αj ◦ αi−1 )(e), 0) = ϕji (e, 0) and is C ∞ , being the restriction of a C ∞ map. The last statement is a direct consequence of the definition of relative topology and Definition 3.2.1.  If M is an n-manifold and B a submanifold of M , the codimension of B in M is defined by codim B = dim M − dim B. Note that open submanifolds are characterized by having codimension zero. In §3.5 methods are developed for proving that various subsets are actually submanifolds, based on the implicit function theorem. For now we do a case “by hand.” 3.2.3 Example. To show that S n ⊂ Rn+1 is a submanifold, it is enough to observe that the charts in the atlas {(Ui± , ψi± )}, i = 1, . . . , n + 1 of S n come from charts of Rn+1 with the submanifold property (see Example 3.1.4): the 2(n + 1) maps n+1 χ± | ±xi > 0 } i : {x ∈ R

→ { y ∈ Rn+1 | (y n+1 + 1)2 > (y 1 )2 + · · · + (y n )2 } given by 1 n+1 χ± ) = (x1 , . . . , xi−1 , xi+1 , . . . , xn+1 , kxk − 1) i (x , . . . , x

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are C ∞ diffeomorphisms, and charts in an atlas of Rn+1 . Since ± 1 n+1 (χ± ) = (x1 , . . . , xi−1 , xi+1 , . . . , xn+1 , 0), i |Ui )(x , . . . , x

they have the submanifold property for S n . Products of Manifolds.



Now we show how to make the product of two manifolds into a manifold.

3.2.4 Definition. Let (S1 , D1 ) and (S2 , D2 ) be two manifolds. The product manifold (S1 × S2 , D1 × D2 ) consists of the set S1 × S2 together with the differentiable structure D1 × D2 generated by the atlas { (U1 × U2 , ϕ1 × ϕ2 ) | (Ui , ϕi ) is a chart of (Si , Di ), i = 1, 2 }. That the set in this definition is an atlas follows from the fact that if ψ1 : U1 ⊂ E1 → V1 ⊂ F1 and ψ2 : U2 ⊂ E2 → V2 ⊂ F2 , then ψ1 × ψ2 is a diffeomorphism iff ψ1 and ψ2 are, and in this case (ψ1 × ψ2 )−1 = ψ1−1 × ψ2−1 . It is clear that the topology on the product manifold is the product topology. Also, if S1 , S2 are finite dimensional, dim(S1 × S2 ) = dim S1 + dim S2 . Inductively, one defines the product of a finite number of manifolds. 3.2.5 Example.

A simple example of a product manifold is the n-torus Tn = S 1 × · · · × S 1 (n times).

Mappings between Manifolds. The following definition introduces two important ideas: the local representative of a map and the concept of a C r map between manifolds. 3.2.6 Definition. Suppose f : M → N is a mapping, where M and N are manifolds. We say f is of class C r (where r is a nonnegative integer), if for each x in M and admissible chart (V, ψ) of N with f (x) ∈ V , there is a chart (U, ϕ) of M satisfying x ∈ U , and f (U ) ⊂ V , and such that the local representative of f , fϕψ = ψ ◦ f ◦ ϕ−1 , is of class C r . (See Figure 3.2.2.)

N

M f

V U

ψ

ϕ

fϕψ

Figure 3.2.2. A local representative of a map

For r = 0, this is consistent with the definition of continuity of f , regarded as a map between topological spaces (with the manifold topologies). 3.2.7 Proposition. Let f : M → N be a continuous map of manifolds. Then f is C r iff the local representatives of f relative to a collection of charts which cover M and N are C r . Proof. Assume that the local representatives of f relative to a collection of charts covering M and N are C r . If (U, ϕ) and (U, ϕ0 ) are charts in M and (V, ψ), (V, ψ 0 ) are charts in N such that fϕψ is C r , then the

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139

composite mapping theorem and condition MA2 of Definition 3.1.1 show that fϕ0 ψ0 = (ψ 0 ◦ ψ −1 ) ◦ fϕψ ◦ (ϕ0 ◦ ϕ−1 )−1 is also C r . Moreover, if ϕ00 and ψ 00 are restrictions of ϕ and ψ to open subsets of U and V , then fϕ00 ψ00 is also C r . Finally, note that if f is C r on open submanifolds of M , then it is C r on their union. That f is C r now follows from the fact that any chart of M can be obtained from the given collection by change of diffeomorphism, restrictions, and/or unions of domains, all three operations preserving the C r character of f . This argument also demonstrates the converse.  Any map from (open subsets of) E to F which is C r in the Banach space sense is C r in the sense of Definition 3.2.6. Other examples of C ∞ maps are the antipodal map x 7→ −x of S n and the translation map by (θ1 , . . . , θn ) on Tn given by (exp(ir1 ), . . . , exp(irn )) 7→ (exp(i(r1 + θ1 )), . . . , exp(i(rn + θn ))). From the previous proposition and the composite mapping theorem, we get the following. 3.2.8 Proposition.

If f : M → N and g : N → P are C r maps, then so is g ◦ f .

3.2.9 Definition. A map f : M → N , where M and N are manifolds, is called a C r diffeomorphism if f is of class C r , is a bijection, and f −1 : N → M is of class C r . If a diffeomorphism exists between two manifolds, they are called diffeomorphic. It follows from Proposition 3.2.8 that the set Diff r (M ) of C r diffeomorphisms of M forms a group under composition. This large and intricate group will be encountered again several times in the book.

Exercises  3.2-1.

Show that

(i) if (U, ϕ) is a chart of M and ψ : ϕ(U ) → V ⊂ F is a diffeomorphism, then (U, ψ ◦ ϕ) is an admissible chart of M , and (ii) admissible local charts are diffeomorphisms.  3.2-2. Show that a C 1 diffeomorphism that is also a C r map is a C r diffeomorphism. Hint: Use the comments after the proof of Theorem 2.5.2.  3.2-3. Show that if Ni ⊂ Mi are submanifolds, i = 1, ..., n, then N1 × · · · × Nn is a submanifold of M1 × · · · × Mn .  3.2-4. Show that every submanifold N of a manifold M is locally closed in M ; that is, every point n ∈ N has a neighborhood U in M such that N ∩ U is closed in U .  3.2-5.

Show that fi : Mi → Ni , i = 1, . . . , n are all C r iff f1 × · · · × fn : M1 × · · · × Mn → N1 × · · · × Nn

is C r .  3.2-6. Let M be a set and {Mi }i∈I a covering of M , each Mi being a manifold. Assume that for every pair of indices (i, j), Mi ∩ Mj is an open submanifold in both Mi and Mj . Show that there is a unique manifold structure on M for which the Mi are open submanifolds. The differentiable structure on M is said to be obtained by the collation of the differentiable structures of Mi .  3.2-7. Show that the map F 7→ F0 = { u ∈ F∗ | u|F = 0 } of G(E) into G(E∗ ) is a C ∞ map. If E = E∗∗ (i.e., E is reflexive) it restricts to a C ∞ diffeomorphism of Gn (E) onto Gn (E∗ ) for all n = 1, 2, . . . . Conclude that RPn is diffeomorphic to Gn (Rn+1 ).

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3. Manifolds and Vector Bundles

 3.2-8. Show that the two differentiable structures of R defined in Example 3.1.8D are diffeomorphic. Hint: Consider the map x 7→ x1/3 .  3.2-9. (i) Show that S 1 and RP1 are diffeomorphic manifolds (see Exercise 3.1-6(b)). (ii) Show that CP1 is diffeomorphic to S 2 (see Exercise 3.1-7(b)).  3.2-10.

Let Mλ = { (x, |x|λ ) | x ∈ R }, where λ ∈ R. Show that

(i) if λ ≤ 0, Mλ is a C ∞ submanifold of R2 ; (ii) if λ > 0 is an even integer, Mλ is a C ∞ submanifold of R2 ; (iii) if λ > 0 is an odd integer or not an integer, then Mλ is a C [λ] submanifold of R2 which is not C [λ]+1 , where [λ] denotes the smallest integer ≥ λ, that is, [λ] ≤ λ < [λ] + 1; (iv) in case (iii), show that Mλ is the union of three disjoint C ∞ submanifolds of R2 .  3.2-11. Let M be a C k manifold. Show that the diagonal ∆ = { (m, m) | m ∈ M } is a closed C k submanifold of M × M .  3.2-12. Let E be a Banach space. Show that the map x 7→ Rx(R2 − kxk2 )−1/2 is a diffeomorphism of the open ball of radius R with E. Conclude that any manifold M modeled on E has an atlas {(Ui , ϕi )} for which ϕi (Ui ) = E.  3.2-13.

If f : M → N is of class C k and S is a submanifold of M , show that f |S is of class C k .

 3.2-14. Let M and N be C r manifolds and f : M → N be a continuous map. Show that f is of class C k , 1 ≤ k ≤ r if and only if for any open set U in N and any C k map g : U → E, E a Banach space, the map g ◦ f : f −1 (U ) → E is C k .  3.2-15. Let π : S n → RPn denote the projection. Show that f : RPn → M is smooth iff the map f ◦ π : S n → M is smooth; here M denotes another smooth manifold.  3.2-16 (Covering Manifolds). Let M and N be smooth manifolds and let p : M → N be a smooth map. The map p is called a covering , or equivalently, M is said to cover N , if p is surjective and each point n ∈ N admits an open neighborhood V such that p−1 (V ) is a union of disjoint open sets, each diffeomorphic via p to V . (i) Path lifting property. Suppose p : M → N is a covering and p(m0 ) = n0 , where n0 ∈ N and m0 ∈ M . Let c : [0, 1] → N be a C k path, k ≥ 0, starting at n0 = c(0). Show that there is a unique C k path d : [0, 1] → M , such that d(0) = m0 and p ◦ d = c. Hint: Partition [0, 1] into a finite set of closed intervals [ti , ti+1 ], i = 0, . . . , n − 1, where t0 = 0 and tn = 1, such that each of the sets c([ti , ti+1 ]) lies entirely in a neighborhood Vi guaranteed by the covering property of p. Let U0 be the open set in the union p−1 (V0 ) containing m0 . Define d0 : [0, t1 ] → U0 by d0 = p−1 ◦ c|[0, t1 ]. Let V1 be the open set containing c([t1 , t2 ]) and U1 be the open set in the union p−1 (V1 ) containing d(t1 ). Define the map d1 : [t1 , t2 ] → U1 by d1 = p−1 ◦ c|[t1 , t2 ]. Now proceed inductively. Show that d so obtained is C k if c is and prove the construction is independent of the partition of [0, 1]. (ii) Homotopy lifting property. In the hypotheses and notations of (i), let H : [0, 1] × [0, 1] → N be a C k map, k ≥ 0 and assume that H(0, 0) = n0 . Show that there is a unique C k -map K : [0, 1] × [0, 1] → M such that K(0, 0) = m0 and p ◦ K = H. Hint: Apply the reasoning in (i) to the square [0, 1] × [0, 1].

3.2 Submanifolds, Products, and Mappings

141

(iii) Show that if two curves in N are homotopic via a homotopy keeping the endpoints fixed, then the lifted curves are also homotopic via a homotopy keeping the endpoints fixed. (iv) Assume that pi : Mi → N are coverings of N with Mi connected, i = 1, 2. Show that if M1 is simply connected, then M1 is also a covering of M2 . Hint: Choose points n0 ∈ N , m1 ∈ M1 , m2 ∈ M2 such that pi (mi ) = n0 , i = 1, 2. Let x ∈ M1 and let c1 (t) be a C k -curve (k is the differentiability class of M1 , M2 , and N ) in M1 such that c1 (0) = m1 , c1 (1) = x. Then c(t) = (p ◦ c1 )(t) is a curve in N connecting n0 to p1 (x). Lift this curve to a curve c2 (t) in M2 connecting m2 to y = c2 (1) and define q : M1 → M2 by q(x) = y. Show by (iii) that q is well defined and C k . Then show that q is a covering. (v) Show that if pi : Mi → N , i = 1, 2 are coverings with M1 and M2 simply connected, then M1 and M2 are C k -diffeomorphic. This is why a simply connected covering of N is called the universal covering manifold of N .  3.2-17 (Construction of the universal covering manifold). Let N be a connected (hence arcwise connected) manifold and fix n0 ∈ N . Let M denote the set of homotopy classes of paths c : [0, 1] → N , c(0) = n0 , keeping the endpoints fixed. Define p : M → N by p([c]) = c(1), where [c] is the homotopy class of c. (i) Show that p is onto since N is arcwise connected. (ii) For an open set U in N define U[c] = { [c ∗ d] | d is a path in U starting at c(1) }. (See Exercise 1.6-6 for the definition of c ∗ d.) Show that B = { ∅, U[c] | c is a path in N starting at n0 and U is open in N } is a basis for a topology on M . Show that if N is Hausdorff, so is M . Show that p is continuous. (iii) Show that M is arcwise connected. Hint: A continuous path ϕ : [0, 1] → M,

ϕ(0) = [c] and ϕ(1) = [d]

is given by ϕ(s) = [cs ], for s ∈ [0, 1/2], and ϕ(s) = [ds ], for s ∈ [1/2, 1], where cs (t) = c((1 − 2s)t),

ds (t) = d((2s − 1)t).

(iv) Show that p is an open map. Hint: If n ∈ p(U[c] ) then the set of points in U that can be joined to n by paths in U is open in N and included in p(U[c] ). (v) Use (iv) to show that p : M → N is a covering. Hint: Let U be a contractible chart domain of N and show that [ p−1 (U ) = U[c] , where the union is over all paths c with p([c]) = n, n a fixed point in U . (vi) Show that M is simply connected. Hint: If ψ : [0, 1] → M is a loop based at [c], that is, ψ is continuous and ψ(0) = ψ(1) = [c], then H : [0, 1] × [0, 1] → M given by H(·, s) = [cs ], cs (t) = c(ts) is a homotopy of [c] with the constant path [c(0)]. (vii) If (U, ϕ) is a chart on N whose domain is such that p−1 (U ) is a disjoint union of open sets in M each diffeomorphic to U (see (v)), define ψ : V → E by ψ = ϕ ◦ p|V . Show that the atlas defined in this way defines a manifold structure on M . Show that M is locally diffeomorphic to N .

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3. Manifolds and Vector Bundles

The Tangent Bundle

Recall that for f : U ⊂ E → V ⊂ F of class C r+1 we define the tangent of f , T f : T U → T V by setting T U = U × E, T V = V × F, and T f (u, e) = (f (u), Df (u) · e) and that the chain rule reads T (g ◦ f ) = T g ◦ T f. If for each open set U in some vector space E, τU : T U → U denotes the projection, the diagram TU

Tf

τU

- TV τV ? - V

? U f

is commutative, that is, f ◦ τU = τV ◦ T f . The tangent operation T can now be extended from this local context to the context of differentiable manifolds and mappings. During the definitions it may be helpful to keep in mind the example of the family of tangent spaces to the sphere S n ⊂ Rn+1 . A major advance in differential geometry occurred when it was realized how to define the tangent space to an abstract manifold independent of any embedding in Rn .4 Several alternative ways to do this can be used according to taste as we shall now list; see Spivak [1979] for further information. Coordinates. Using transformation properties of vectors under coordinate changes, one defines a tangent vector at m ∈ M to be an equivalence class of triples (U, ϕ, e), where ϕ : U → E is a chart and e ∈ E, with two triples identified if they are related by the tangent of the corresponding overlap map evaluated at the point corresponding to m ∈ M . Derivations. This approach characterizes a vector by specifying a map that gives the derivative of a sufficiently smooth function in the direction of that vector. (0)

(1)

(j)

Ideals. This is a variation of alternative 2. Here Tm M is defined to be the dual of Im /Im , where Im is the ideal of functions on M vanishing up to order j at m. Curves. This is the method followed here. We abstract the idea that a tangent vector to a surface is the velocity vector of a curve in the surface. If [a, b] is a closed interval, a continuous map c : [a, b] → M is said to be differentiable at the endpoint a if there is a chart (U, ϕ) at c(a) such that lim t↓a

(ϕ ◦ c)(t) − (ϕ ◦ c)(a) t−a

exists and is finite; this limit is denoted by (ϕ ◦ c)0 (a). If (V, ψ) is another chart at c(a) and we let v = (ϕ ◦ c)(t) − (ϕ ◦ c)(a), then in U ∩ V we have (ψ ◦ ϕ−1 )((ϕ ◦ c)(t)) − (ψ ◦ ϕ−1 )((ϕ ◦ c)(a)) = D(ψ ◦ ϕ−1 )((ϕ ◦ c)(a)) · v + o(kvk), 4 The history is not completely clear to us, but (as was mentioned in the historical introduction, this idea seems to be primarily due to Riemann, Weyl, and, in addition, Levi-Civit` a played an important role, and was “well known” by 1920.

3.3 The Tangent Bundle

whence

143

(ψ ◦ c)(t) − (ψ ◦ c)(a) D(ψ ◦ ϕ−1 )(ϕ ◦ c)(a) · v o(kvk) = + . t−a t−a t−a

Since lim t↓a

it follows that lim t↓a

v = (ϕ ◦ c)0 (a) t−a

and

lim t↓a

o(kvk) =0 t−a

[(ψ ◦ c)(t) − (ψ ◦ c)(a)] = D(ψ ◦ ϕ−1 )((ϕ ◦ c)(a)) · (ϕ ◦ c)0 (a) t−a

and therefore the map c : [a, b] → M is differentiable at a in the chart (U, ϕ) iff it is differentiable at a in the chart (V, ψ). In summary, it makes sense to speak of differentiability of curves at an endpoint of a closed interval . The map c : [a, b] → M is said to be differentiable if c|]a, b[ is differentiable and if c is differentiable at the endpoints a and b. The map c : [a, b] → M is said to be of class C 1 if it is differentiable and if (ϕ ◦ c)0 : [a, b] → E is continuous for any chart (U, ϕ) satisfying U ∩ c([a, b]) 6= ∅, where E is the model space of M . 3.3.1 Definition. Let M be a manifold and m ∈ M . A curve at m is a C 1 map c : I → M from an interval I ⊂ R into M with 0 ∈ I and c(0) = m. Let c1 and c2 be curves at m and (U, ϕ) an admissible chart with m ∈ U . Then we say c1 and c2 are tangent at m with respect to ϕ if and only if (ϕ◦c1 )0 (0) = (ϕ◦c2 )0 (0). Thus, two curves are tangent with respect to ϕ if they have identical tangent vectors (same direction and speed) in the chart ϕ; see Figure 3.3.1.

I

U

0

m

Figure 3.3.1. Tangent curves

The reader can safely assume in what follows that I is an open interval; the use of closed intervals becomes essential when defining tangent vectors to a manifold with boundary at a boundary point; this will be discussed in Chapter 7. 3.3.2 Proposition. Let c1 and c2 be two curves at m ∈ M . Suppose (Uβ , ϕβ ) are admissible charts with m ∈ Uβ , β = 1, 2. Then c1 and c2 are tangent at m with respect to ϕ1 if and only if they are tangent at m with respect to ϕ2 . Proof. By taking restrictions if necessary we may suppose that U1 = U2 . Since we have the identity 1 0 ϕ2 ◦ci = (ϕ2 ◦ϕ−1 1 )◦(ϕ1 ◦ci ), the C composite mapping theorem in Banach spaces implies that (ϕ2 ◦c1 ) (0) = 0 0 0 (ϕ2 ◦ c2 ) (0) iff (ϕ1 ◦ c1 ) (0) = (ϕ1 ◦ c2 ) (0).  This proposition guarantees that the tangency of curves at m ∈ M is a notion that is independent of the chart used . Thus we say c1 , c2 are tangent at m ∈ M if c1 , c2 are tangent at m with respect to ϕ, for any local chart ϕ at m. It is evident that tangency at m ∈ M is an equivalence relation among curves at m. An equivalence class of such curves is denoted [c]m , where c is a representative of the class.

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3.3.3 Definition. For a manifold M and m ∈ M the tangent space to M at m is the set of equivalence classes of curves at m: Tm M = { [c]m | c is a curve at m }. For a subset A ⊂ M , let T M |A =

[

Tm M

(disjoint union).

m∈A

We call T M = T M |M the tangent bundle of M . The mapping τM : T M → M defined by τM ([c]m ) = m is the tangent bundle projection of M . Let us show that if M = U , an open set in a Banach space E, T U as defined here can be identified with U × E. This will establish consistency with our usage of T in §2.3. 3.3.4 Lemma. Let U be an open subset of E, and c be a curve at u ∈ U . Then there is a unique e ∈ E such that the curve cu,e defined by cu,e (t) = u + te (with t belonging to an interval I such that cu,e (I) ⊂ U ) is tangent to c at u. Proof. By definition, Dc(0) is the unique linear map in L(R, E) such that the curve g : R → E given by g(t) = u + Dc(0) · t is tangent to c at t = 0. If e = Dc(0) · 1, then g = cu,e .  Define a map i : U × E → T U by i(u, e) = [cu,e ]u . The preceding lemma says that i is a bijection and thus we can define a manifold structure on T U by means of i. The tangent space Tm M at each point m ∈ M has an intrinsic vector space structure. This vector space structure can be defined directly by showing that addition and scalar multiplication can be defined by the corresponding operations in charts and that this definition is independent of the chart. This idea is very important in the general study of vector bundles and we shall return to this point in §3.4. We will use this basic fact in the discussion that follows. We will denote vectors tangent to M at the point m ∈ M by notation such as vm . Thus, by construction, such a tangent vector is an equivalence class of curves, and so we can write vm = [c]m for some curve c at m. Tangents of Mappings. It will be convenient to define the tangent of a mapping before showing that T M is a manifold. The idea is simply that the derivative of a map can be characterized by its effect on tangents to curves. 3.3.5 Lemma. Suppose c1 and c2 are curves at m ∈ M and are tangent at m. Let f : M → N be of class C 1 . Then f ◦ c1 and f ◦ c2 are tangent at f (m) ∈ N . Proof. From the C 1 composite mapping theorem and the remarks prior to Definition 3.3.1, it follows that f ◦ c1 and f ◦ c2 are of class C 1 . For tangency, let (V, ψ) be a chart on N with f (m) ∈ V . We must show that (ψ ◦ f ◦ c1 )0 (0) = (ψ ◦ f ◦ c2 )0 (0). But ψ ◦ f ◦ cα = (ψ ◦ f ◦ ϕ−1 ) ◦ (ϕ ◦ cα ), where (U, ϕ) is a chart on M with f (U ) ⊂ V . Hence the result follows from the C 1 composite mapping theorem.  Now we are ready to consider the intrinsic way to look at the derivative. 3.3.6 Definition.

If f : M → N is of class C 1 , we define T f : T M → T N by T f ([c]m ) = [f ◦ c]f (m) .

We call T f the tangent of f . The map T f is well defined, for if we choose any other representative from [c]m , say c1 , then c and c1 are tangent at m and hence f ◦ c and f ◦ c1 are tangent at f (m), that is, [f ◦ c]f (m) = [f ◦ c1 ]f (m) . By construction the following diagram commutes.

3.3 The Tangent Bundle

TM

Tf

145

- TN

τM

τN ? - N

? M f

This diagram says something very fundamental about the tangent map, namely that T f takes a tangent vector at the point m ∈ M to a tangent vector at the corresponding point f (m) ∈ N . We denote the restriction of T f to the tangent space at m by Tm f . Thus, Tm f : Tm M → Tf (m) N . Tmf vm m f

M

N

Figure 3.3.2. The tangent of a map f takes a tangent vector vm at the point m to a vector at the point f (m).

Some additional basic properties of the tangent operation are summarized in the following. 3.3.7 Theorem (Composite Mapping Theorem). (i) Suppose f : M → N and g : N → K are C r maps of manifolds. Then g ◦ f : M → K is of class C r and T (g ◦ f ) = T g ◦ T f. (ii) If h : M → M is the identity map, then T h : T M → T M is the identity map. (iii) If f : M → N is a diffeomorphism, then T f : T M → T N is a bijection and (T f )−1 = T (f −1 ). Proof. (i) Let (U, ϕ), (V, ψ), (W, ρ) be charts of M, N, K, with f (U ) ⊂ V and g(V ) ⊂ W . Then the local representatives are (g ◦ f )ϕρ = ρ ◦ g ◦ f ◦ ϕ−1 = ρ ◦ g ◦ ψ −1 ◦ ψ ◦ f ◦ ϕ−1 = gψρ ◦ fϕψ . By the composite mapping theorem in Banach spaces, this, and hence g ◦ f , is of class C r . Moreover, T (g ◦ f )[c]m = [g ◦ f ◦ c](g◦f )(m) and (T g ◦ T f )[c]m = T g([f ◦ c]f (m) ) = [g ◦ f ◦ c](g◦f )(m) . Hence T (g ◦ f ) = T g ◦ T f . Part (ii) follows from the definition of T . For (iii), f and f −1 are diffeomorphisms with f ◦ f −1 the identity of N , while f −1 ◦ f is the identity on M . Using (i) and (ii), T f ◦ T f −1 is the identity on T N while T f −1 ◦ T f is the identity on T M . Thus (iii) follows. 

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Next, let us show that in the case of local manifolds, T f as defined in §2.4, which we temporarily denote f 0 , coincides with T f as defined here. 3.3.8 Lemma. Let U ⊂ E and V ⊂ F be local manifolds (open subsets) and f : U → V be of class C 1 . Let i : U × E → T U be the map defined following Lemma 3.3.4. Then the diagram U ×E

f0

i

- V ×F i ? - TV

? TU Tf commutes; that is, T f ◦ i = i ◦ f 0 .

Proof. For (u, e) ∈ U × E, we have (T f ◦ i)(u, e) = T f · [cu,e ]u = [f ◦ cu,e ]f (u) . Also, we have the identities (i ◦ f 0 )(u, e) = i(f (u), Df (u) · e) = [cf (u),D f (u)·e ]f (u) . These will be equal provided the curves t 7→ f (u + te) and t 7→ f (u) + t(Df (u) · e) are tangent at t = 0. But this is clear from the definition of the derivative and the composite mapping theorem.  This lemma states that if we identify U × E and T U by means of i then we should correspondingly identify f 0 and T f . Thus we will just write T f and suppress the identification. Thus, in local charts, we simply identify T f with the map (u, e) 7→ (f (u), Df (u) · e). Holding u fixed, this map is of course linear. This reflects an important, basic, and intrinsic fact about the tangent map, namely that for each m ∈ M , the map Tm f : Tm M → Tf (m) M is linear. Theorem 3.3.7 implies the following. 3.3.9 Lemma. If f : U ⊂ E → V ⊂ F is a C r diffeomorphism, then T f : U × E → V × F is a C r−1 diffeomorphism. The Manifold Structure on T M . For a chart (U, ϕ) on a manifold M , we define T ϕ : T U → T (ϕ(U )) by T ϕ([c]u ) = (ϕ(u), (ϕ ◦ c)0 (0)). Then T ϕ is a bijection, since ϕ is a diffeomorphism. Hence, on T M we can regard (T U, T ϕ) as a local chart. 3.3.10 Theorem. Let M be a C r+1 manifold and A an atlas of admissible charts. Then T A = { (T U, T ϕ) | (U, ϕ) ∈ A } is a C r atlas of T M called the natural atlas. Proof. Since the union of chart domains of A is M , the union of the corresponding T U is T M . To verify MA2, suppose we have T Ui ∩ T Uj 6= ∅. Then Ui ∩ Uj 6= ∅ and therefore the overlap map ϕi ◦ ϕ−1 can be j −1 −1 formed by restriction of ϕi ◦ ϕ−1 to ϕ (U ∩ U ). The chart overlap map T ϕ ◦ (T ϕ ) = T (ϕ ◦ ϕ j i j i j i j j ) is a C r diffeomorphism by Lemma 3.3.9.  Hence T M has a natural C r manifold structure induced by the differentiable structure of M . If M is ndimensional, Hausdorff, and second countable, T M will be 2n-dimensional, Hausdorff, and second countable. Since the local representative of τM is (ϕ ◦ τM ◦ T ϕ−1 )(u, e) = u, the tangent bundle projection is a C r map. Let us next develop some of the simplest properties of tangent maps. First of all, let us check that tangent maps are smooth. 3.3.11 Proposition. Let M and N be C r+1 manifolds, and let f : M → N be a map of class C r+1 . Then T f : T M → T N is a map of class C r . Proof. It is enough to check that T f is a C r map using the natural atlas. For m ∈ M choose charts (U, ϕ) and (V, ψ) on M and N so that m ∈ U , f (m) ∈ V and fϕψ = ψ ◦ f ◦ ϕ−1 is of class C r+1 . Using (T U, T ϕ) for T M and (T V, T ψ) for T N , the local representative (T f )T ϕ,T ψ = T ψ ◦ T f ◦ T ϕ−1 = T fϕψ is given by T fϕψ (u, e) = (u, Dfϕψ (u) · e), which is a C r map. 

3.3 The Tangent Bundle

147

Higher Order Tangents. Now that T M has a manifold structure we can form higher tangents. For mappings f : M → N of class C r , define T r f : T r M → T r N inductively to be the tangent of T r−1 f : T r−1 M → T r−1 N . Induction shows: If f : M → N and g : N → K are C r mappings of manifolds, then g ◦ f is of class C r and T r (g ◦ f ) = T r g ◦ T r f . Let us apply the tangent construction to the manifold T M and its projection. This gives the tangent bundle of T M , namely τT M : T (T M ) → T M . In coordinates, if (U, ϕ) is a chart in M , then (T U, T ϕ) is a chart of T M , (T (T U ), T (T ϕ)) is a chart of T (T M ), and thus the local representative of τT M is (T ϕ ◦ τT M ◦ T (T ϕ−1 )) : (u, e, e1 , e2 ) 7→ (u, e). On the other hand, taking the tangent of the map τM : T M → M , we get T τM : T (T M ) → T M . The local representative of T τM is (T ϕ ◦ T τM ◦ T (T ϕ−1 ))(u, e, e1 , e2 ) = T (ϕ ◦ τM ◦ T ϕ−1 )(u, e, e1 , e2 ) = (u, e1 ). Applying the commutative diagram for T f following Definition 3.3.6 to the case f = τM , we get what is commonly known as the dual tangent rhombic:

τT M

T (T M ) @

@ T τM @ R @ TM

TM @ @ τM

τM @ R @

M



Tangent Bundles of Product Manifolds. denoted by single letters such as v ∈ Tm M . 3.3.12 Proposition. projections. The map

Here and in what follows, tangent vectors will often be

Let M1 and M2 be manifolds and pi : M1 × M2 → Mi , i = 1, 2, the two canonical (T p1 , T p2 ) : T (M1 × M2 ) → T M1 × T M2

defined by (T p1 , T p2 )(v) = (T p1 (v), T p2 (v)) is a diffeomorphism of the tangent bundle T (M1 × M2 ) with the product manifold T M1 × T M2 . Proof.

The local representative of this map is (u1 , u2 , e1 , e2 ) ∈ U1 × U2 × E1 × E2 7→ ((u1 , e1 ) , (u2 , e2 )) ∈ (U1 × E1 ) × (U2 × E2 ) ,

which clearly is a local diffeomorphism.



Partial Tangents. Since the tangent is just a global version of the derivative, statements concerning partial derivatives might be expected to have analogues on manifolds. To effect these analogies, we globalize the definition of partial derivatives. Let M1 , M2 , and N be manifolds, and f : M1 × M2 → N be a C r map. For (p, q) ∈ M1 × M2 , let ip : M2 → M1 × M2 and iq : M1 → M1 × M2 be given by ip (y) = (p, y),

iq (x) = (x, q),

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3. Manifolds and Vector Bundles

and define T1 f (p, q) : Tp M1 → Tf (p,q) N and T2 f(p,q) : Tq M2 → Tf (p,q) N by T1 f (p, q) = Tp (f ◦ iq ),

T2 f (p, q) = Tq (f ◦ ip ).

With these notations the following proposition giving the behavior of T under products is a straightforward verification using the definition and local differential calculus. In the following proposition we will again use the important fact that each tangent space Tm M to a manifold at m ∈ M , has a natural vector space structure consistent with the vector space structure in local charts. 3.3.13 Proposition. Let M1 , M2 , N , and P be manifolds, gi : P → Mi , i = 1, 2, and f : M1 × M2 → N be C r maps, r ≥ 1. Identify T (M1 × M2 ) with T M1 × T M2 . Then the following statements hold. (i) T (g1 × g2 ) = T g1 × T g2 . (ii) T f (up , vq ) = T1 f (p, q)(up ) + T2 f (p, q)(vq ), for up ∈ Tp M1 and vq ∈ Tq M2 . (iii) (Implicit Function Theorem.) If T2 f (p, q) is an isomorphism, then there exist open neighborhoods U of p in M1 , W of f (p, q) in N , and a unique C r map g : U ×W → M2 such that for all (x, w) ∈ U ×W , f (x, g(x, w)) = w. In addition, T1 g(x, w) = −(T2 f (x, g(x, w)))−1 ◦ (T1 f (x, g(x, w))) and T2 g(x, w) = (T2 f (x, g(x, w)))−1 . 3.3.14 Examples. A. The tangent bundle T S 1 of the circle. We claim that T S 1 , the tangent bundle of the circle is diffeomorphic to S 1 × R. As we shall mention in Example C, it is not true that the tangent bundle of the sphere is the product of the sphere with R2 . See Figure 3.3.3. y

R

s λ x s

λ Trivial tangent bundle

p Nontrivial tangent bundle

Figure 3.3.3. Trivial and nontrivial tangent bundles.

3.3 The Tangent Bundle

149

As in the figure, a diffeomorphism between T S 1 and S 1 × R is given by taking a tangent vector to the circle and mapping it to the base point of the tangent vector together with the oriented length of the tangent vector. The oriented length is the length of the tangent vector taken in, say the counterclockwise sense of going around the circle. Notice that one does not have such a notion for the two sphere. This fact can also be seen more formally using charts as follows. Consider the atlas with the four charts { (Ui± , ψi± ) | i = 1, 2 } of S 1 = { (x, y) ∈ R2 | x2 + y 2 = 1 } from Example 3.1.4. Let us construct the natural atlas for T S 1 = { ((x, y), (u, v)) ∈ R2 × R2 | x2 + y 2 = 1, h(x, y), (u, v)i = 0 }. Since the map ψ1+ : U1+ = { (x, y) ∈ S 1 | x > 0 } → ]−1, 1[ is given by ψ1+ (x, y) = y, by definition of the tangent we have T(x,y) ψ1+ (u, v) = (y, v),

T ψ1+ : T U1+ → ]−1, 1[×R.

Proceed in the same way with the other three charts. Thus, for example, T(x,y) ψ2−1 (u, v) = (x, u) and hence for x ∈ ]−1, 0[, ! p yv − + −1 2 1 − y , −p . (T ψ2 ◦ T (ψ1 ) )(y, v) = 1 − y2 This gives a complete description of the tangent bundle. But more can be said. Thinking of S 1 as the multiplicative group of complex numbers with modulus 1, we shall show that the group operations are C ∞ : the inversion I : s 7→ s−1 has local representative (ψ1± ◦ I ◦ (ψ1± )−1 )(x) = −x and the composition C : (s1 , s2 ) 7→ s1 s2 has local representative q q (ψ1 ◦ C ◦ (ψ1± × ψ1± )−1 )(x1 , x2 ) = x1 1 − x22 + x2 1 − x21 (here ± can be taken in any order). Thus for each s ∈ S 1 , the map Ls : S 1 → S 1 defined by Ls (s0 ) = ss0 , is a diffeomorphism. This enables us to define a map λ : T S 1 → S 1 × R by λ(vs ) = (s, Ts L−1 s (vs )), which is easily seen to be a diffeomorphism. B. The tangent bundle T Tn to the n-torus. it follows that T Tn ∼ = Tn × Rn .

Since Tn = S 1 × · · · × S 1 (n times) and T S 1 ∼ = S 1 × R,

C. The tangent bundle T S 2 to the sphere. The previous examples yielded trivial tangent bundles. In general this is not the case, the tangent bundle to the two-sphere being a case in point, which we now describe. Choose the atlas with six charts { (Ui± , ψi± ) | i = 1, 2, 3 } of S 2 that were given in Example 3.1.4. Since ψ1± : U1+ = { (x1 , x2 , x3 ) ∈ S 2 | x1 > 0 } → D1 (0) = { (x, y) ∈ R2 | x2 + y 2 < 1 }, ψ1+ (x1 , x2 , x3 ) = (x2 , x3 ), we have T(x1 ,x2 ,x3 ) ψ1+ (v 1 , v 2 , v 3 ) = (x2 , x3 , v 2 , v 3 ), where x1 v 1 + x2 v 2 + x3 v 3 = 0. Similarly, construct the other five charts. For example, one of the twelve overlap maps for x2 + y 2 < 1, and y < 0, is (T ψ3− ◦ (T ψ1+ )−1 )(x, y, u, v) =

p

1−

x2

−

y 2 , x,

−ux

vy

!

p −p ,u . 1 − x2 − y 2 1 − x2 − y 2

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3. Manifolds and Vector Bundles

One way to see that T S 2 is not trivial is to use the topological fact that any vector field on S 2 must vanish somewhere. We shall prove this fact in §7.5. 

Exercises  3.3-1.

Let M and N be manifolds and f : M → N .

(i) Show that (a) if f is C ∞ , then graph (f ) = { (m, f (m)) ∈ M × N | m ∈ M } is a C ∞ submanifold of M × N and ∼ T(m,f (m)) (graph(f )) ⊕ Tf (m) N for all m ∈ M . (b) T(m,f (m)) (M × N ) = (c) Show that the converse of (a) is false. Hint: x ∈ R 7→ x1/3 ∈ R. (d) Show that if (a) and (b) hold, then f is C ∞ . (ii) If f is C ∞ show that the canonical projection of graph(f ) onto M is a diffeomorphism. (iii) Show that T(m,f (m)) (graph(f )) ∼ = graph(Tm f ) = { (vm , Tm f (vm )) | vm ∈ Tm M } ⊂ Tm M × Tf (m) N.  3.3-2. (i) Show that there is a map sM : T (T M ) → T (T M ) such that sM ◦ sM = identity and the diagram

T (T M )

sM



sM

@ @ τT M

- T (T M ) T τM

@ R @ TM commutes. Hint: In a chart, sM (u, e, e1 , e2 ) = (u, e1 , e, e2 ).) One calls sM the canonical involution on M and says that T (T M ) is a symmetric rhombic. (ii) Verify that for f : M → N of class C 2 , T 2 f ◦ sM = sN ◦ T 2 f . (iii) If X is a vector field on M , that is, a section of τM : T M → M , show that T X is a section of T τM : T 2 M → T M and X 1 = sM ◦ T X is a section of τT M : T 2 M → T M . (A section σ of a map f : A → B is a map σ : B → A such that f ◦ σ = identity on B.)  3.3-3. (i) Let S(S 2 ) = { (v) ∈ T S 2 | k (v) k = 1 } be the circle bundle of S 2 . Prove that S(S 2 ) is a submanifold of T S 2 of dimension three. (ii) Define f : S(S 2 ) → RP3 by f (x, y, (v)) = the line through the origin in R4 determined by the vector with components (x, y, v 1 , v 2 ). Show that f is a diffeomorphism.

3.4 Vector Bundles

151

 3.3-4. Let M be an n-dimensional submanifold of RN . Define the Gauss map Γ : M → Gn,N −n by Γ(m) = Tm M − m, that is, Γ(m) is the n-dimensional subspace of RN through the origin, which, when translated by m, equals Tm M . Show that Γ is a smooth map.  3.3-5. Let f : T2 → R be a smooth map. Show that f has at least four critical points (points where T f vanishes). Hint: Parametrize T2 using angles θ, ϕ and locate the maximum and minimum points of f (θ, ϕ) for ϕ fixed, say (θmax (ϕ), ϕ) and (θmin (ϕ), ϕ); now maximize and minimize f as ϕ varies. How many critical points must f : S 2 → R have?

3.4

Vector Bundles

Roughly speaking, a vector bundle is a manifold with a vector space attached to each point. During the formal definitions we may keep in mind the example of the tangent bundle to a manifold, such as the 2-sphere S 2 . Similarly, the collection of normal lines to S 2 ⊂ R3 form a vector bundle. Definition of a Vector Bundle. The definitions will follow the pattern of those for a manifold. Namely, we obtain a vector bundle by smoothly patching together local vector bundles. The following terminology for vector space products and maps will be useful. 3.4.1 Definition. Let E and F be Banach spaces with U an open subset of E. We call the Cartesian product U × F a local vector bundle. We call U the base space, which can be identified with U × {0}, the zero section. For u ∈ U , {u} × F is called the fiber of u, which we endow with the vector space structure of F. The map π : U × F → U given by π(u, f ) = u is called the projection of U × F. (Thus, the fiber over u ∈ U is π −1 (u). Also note that U × F is an open subset of E × F and so is a local manifold.) Next, we introduce the idea of a local vector bundle map. The main idea is that such a map must send a fiber linearly to a fiber. 3.4.2 Definition. Let U × F and U 0 × F0 be local vector bundles. A map ϕ : U × F → U 0 × F0 is called a C r local vector bundle map if it has the form ϕ(u, f ) = (ϕ1 (u), ϕ2 (u) · f ) where ϕ1 : U → U 0 and ϕ2 : U → L(F, F0 ) are C r . A local vector bundle map that has an inverse which is also a local vector bundle map is called a local vector bundle isomorphism. (See Figure 3.4.1.) F

F

U

U

Figure 3.4.1. A vector bundle

A local vector bundle map ϕ : U × F → U 0 × F0 maps the fiber {u} × F into the fiber {ϕ1 (u)} × F0 and so restricted is linear. By Banach’s isomorphism theorem it follows that a local vector bundle map ϕ with ϕ1 a local diffeomorphism is a local vector bundle isomorphism iff ϕ2 (u) is a Banach space isomorphism for every u ∈ U .

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3. Manifolds and Vector Bundles

Supplement 3.4A Smoothness of Local Vector Bundle Maps In some examples, to check whether a map ϕ is a C ∞ local vector bundle map, one is faced with the rather unpleasant task of verifying that ϕ2 : U → L(F, F0 ) is C ∞ . It would be nice to know that the smoothness of ϕ as a function of two variables suffices. This is the context of the next proposition. We state the result for C ∞ , but the proof contains a C r result (with an interesting derivative loss) which is discussed in the ensuing Remark A. 3.4.3 Proposition. A map ϕ : U × F → U 0 × F0 is a C ∞ local vector bundle map iff ϕ is C ∞ and is of the form ϕ(u, f ) = (ϕ1 (u), ϕ2 (u) · f ), where ϕ1 : U → U 0 and ϕ2 : U → L(F, F0 ). Proof (Craioveanu and Ratiu [1976]). The evaluation map ev : L(F, F0 ) × F → F0 ; ev(T, f ) = T (f ) is clearly bilinear and continuous. First assume ϕ is a C r local vector bundle map, so ϕ1 : U → U 0 and ϕ2 : U → L(F, F0 ) are C r . Now write ϕ2 (u) · f = (ev ◦ (ϕ2 × I))(u, f ). By the composite mapping theorem, it follows that ϕ2 is C r as a function of two variables. Thus ϕ is a C r map by Proposition 2.4.12(iii). Conversely, assume ϕ(u, f ) = (ϕ1 (u), ϕ2 (u) · f ) is C ∞ . Then again by Proposition 2.4.12(iii), ϕ1 (u) and ϕ2 (u) · f are C ∞ as functions of two variables. To show that ϕ2 : U → L(E, F0 ) is C ∞ , it suffices to prove the following: if h : U × F → F0 is C r , r ≥ 1, and such that h(u, ·) ∈ L(F, F0 ) for all u ∈ U , then the map h0 : U → L(F, F0 ), defined by h0 (u) = h(u, ·) is C r−1 . This will be shown by induction on r. If r = 1 we prove continuity of h0 in a disk around u0 ∈ U in the following way. By continuity of Dh, there exists ε > 0 such that for all u ∈ Dε (u0 ) and v ∈ Dε (0), kD1 h(u, v)k ≤ N for some N > 0. The mean value inequality yields kh(u, v) − h(u0 , v)k ≤ N ku − u0 k for all u, u0 ∈ Dε (u0 ) and v ∈ Dε (0). Thus kh0 (u) − h0 (u0 )k = sup kh(u, v) − h(u0 , v)k < kvk≤1

N ku − u0 k, ε

proving that h0 is continuous. Let r > 1 and inductively assume that the statement is true for r − 1. Let S : L(F, L(E, F0 )) → L(E, L(F, F0 )) be the canonical isometry: S(T )(e) · f = T (f ) · e. We shall prove that Dh0 = S ◦ (D1 h)0 ,

(3.4.1)

where (D1 h)0 (u) · v = D1 h(u, v). Thus, if h is C r , D1 h is C r−1 , by induction (D1 h)0 is C r−2 , and hence by equation (3.4.1), Dh0 will be C r−2 . This will show that h0 is C r−1 . For equation (3.4.1) to make sense, we first show that D1 h(u, ·) ∈ L(F, L(E, F0 )). Since

lim [h0 (u + tw) − h0 (u)] · v

D1 h(u, v) · w =

t→0

t

= lim An v, n→∞

for all v ∈ F, where     1 An = n h0 u + w − h0 (u) ∈ L(F, F0 ), n

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153

it follows by the uniform boundedness principle (or rather its Corollary 2.2.21) that D1 h(u, ·) · w ∈ L(F, F0 ). Thus (v, w) 7→ D1 h(u, v) · w is linear continuous in each argument and hence is bilinear continuous (Exercise 2.2-10), and consequently v 7→ D1 h(u, v) ∈ L(E, F0 ) is linear and continuous. Relation (3.4.1) is proved in the following way. Fix u0 ∈ U and let ε and N be positive constants such that kD1 h(u, v) − D1 h(u0 , v)k ≤ N ku − u0 k (3.4.2) for all u, u0 ∈ D2ε (u0 ) and v ∈ Dε (0). Apply the mean value inequality to the C r−1 map g(u) = h(u, v) − D1 h(u0 , v) · u for fixed u0 ∈ D2ε (u0 ) and v ∈ Dε (0) to get kh(u + w, v) − h(u, v) − D1 h(u0 , v) · wk = kg(u + w) − g(u)k ≤ kwk sup kDg(u + tw)k t∈[0,1]

= kwk sup kD1 h(u + tw, v) − D1 h(u0 , v)k t∈[0,1]

for w ∈ Dε (u0 ). Letting u0 → u and taking into account equation (3.4.2) we get kh(u + w, v) − h(u, v) − D1 h(u, v) · wk ≤ N kwk2 ; that is, kh0 (u + w) · v − h0 (u) · v − [(S ◦ (D1 h)0 )(u) · w](v)k ≤ N kwk2 for all v ∈ Dε (0), and hence kh0 (u + w) − h0 (u) − (S ◦ (D1 h)0 ) · wk ≤ thus proving equation (3.4.1).

N kwk2 ε 

Remarks A. If F is finite dimensional and if h : U × F → F0 is C r , r ≥ 1, and is such that h(u, ·) ∈ L(F, F0 ) for all u ∈ U , then h0 : U → L(F, F0 ) given by h0 (u) = h(u, ·) is also C r . In other words, Proposition 3.4.3 holds for C r -maps. Indeed, since F = Rn for some n, L(F, F0 ) ∼ = F0 × · · · × F0 (n times) so it suffices to prove the statement for F = R. Thus we want to show that if h : U × R → F0 is C r and h(u, 1) = g(u) ∈ F0 , then g : U → F0 is also C r . Since h(u, x) = xg(u) for all (u, x) ∈ U × R by linearity of h in the second argument, it follows that h0 = g is a C r map. B. If F is infinite dimensional the result in the proof of Proposition 3.4.3 cannot be improved even if r = 0. The following counterexample is due to A.J. Tromba. Let h : [0, 1] × L2 [0, 1] → L2 [0, 1] be given by   Z 1 2πt ϕ(t) dt h(x, ϕ) = sin x 0 if x 6= 0, and h(0, ϕ) = 0. Continuity at each x 6= 0 is obvious and at x = 0 it follows by the Riemann– Lebesgue lemma (the Fourier coefficients of a uniformly bounded sequence in L2 relative to an orthonormal set converge to zero). Thus h is C 0 . However, since    2πt 1 x 4π h x, sin = − sin , x 2 4π x

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√ √ we have h(1/n, sin 2πnt) = 1/2 and therefore its L2 -norm is 1/ 2; this says that kh0 (1/n)k ≥ 1/ 2 and thus h0 is not continuous. 

Any linear map A ∈ L(E, F) defines a local vector bundle map ϕA : E × E → E × F by ϕ(u, e) = (u, Ae). Another example of a local vector bundle map was encountered in §2.4: if the map f : U ⊂ E → V ⊂ F is C r+1 , then T f : U × E → V × F is a C r local vector bundle map and T f (u, e) = (f (u), Df (u) · e). Using these local notions, we are now ready to define a vector bundle. 3.4.4 Definition. Let S be a set. A local bundle chart of S is a pair (W, ϕ) where W ⊂ S and ϕ : W ⊂ S → U × F is a bijection onto a local bundle U × F; U and F may depend on ϕ. A vector bundle atlas on S is a family B = {(Wi , ϕi )} of local bundle charts satisfying: VB1. = MA1 of Definition 3.1.1: B covers S; and VB2. for any two local bundle charts (Wi , ϕi ) and (Wj , ϕj ) in B with Wi ∩ Wj 6= ∅, ϕi (Wi ∩ Wj ) is a local vector bundle, and the overlap map ψji = ϕj ◦ ϕ−1 restricted to ϕi (Wi ∩ Wj ) is a C ∞ i local vector bundle isomorphism. If B1 and B2 are two vector bundle atlases on S, we say that they are VB-equivalent if B1 ∪ B2 is a vector bundle atlas. A vector bundle structure on S is an equivalence class of vector bundle atlases. A vector bundle E is a pair (S, V), where S is a set and V is a vector bundle structure on S. A chart in an atlas of V is called an admissible vector bundle chart of E. As with manifolds, we often identify E with the underlying set S. The intuition behind this definition is depicted in Figure 3.4.2. As in the case of manifolds, if we make a choice of vector bundle atlas B on S then we obtain a maximal vector bundle atlas by including all charts whose overlap maps with those in B are C ∞ local vector bundle isomorphisms. Hence a particular vector bundle atlas suffices to specify a vector bundle structure on S. Vector bundles are special types of manifolds. Indeed VB1 and VB2 give MA1 and MA2 in particular, so V induces a differentiable structure on S. 3.4.5 Definition.

For a vector bundle E = (S, V) we define the zero section (or base) by  B = e ∈ E | there exists (W, ϕ) ∈ V and u ∈ U with e = ϕ−1 (u, 0) ,

that is, B is the union of all the zero sections of the local vector bundles (identifying W with a local vector bundle via ϕ : W → U × F). If (U, ϕ) ∈ V is a vector bundle chart, and b ∈ U with ϕ(b) = (u, 0), let Eb,ϕ denote the subset ϕ−1 ({u}×F) of S together with the structure of a vector space induced by the bijection ϕ. The next few propositions derive basic properties of vector bundles that are sometimes included in the definition. 3.4.6 Proposition.

(i) If b lies in the domain of two local bundle charts ϕ1 and ϕ2 , then Eb,ϕ1 = Eb,ϕ2 ,

where the equality means equality as topological spaces and as vector spaces.

3.4 Vector Bundles

155

E B

W

Figure 3.4.2. Vector bundle charts

(ii) For v ∈ E, there is a unique b ∈ B such that v ∈ Eb,ϕ , for some (and so all ) (U, ϕ). (iii) B is a submanifold of E. (iv) The map π, defined by π : E → B, π(e) = b [in (ii)] is surjective and C ∞ . Proof. (i) Suppose ϕ1 (b) = (u1 , 0) and ϕ2 (b) = (u2 , 0). We may assume that the domains of ϕ1 and ϕ2 are identical, for Eb,ϕ is unchanged if we restrict ϕ to any local bundle chart containing b. Then α = ϕ1 ◦ϕ−1 2 is a local vector bundle isomorphism. But we have −1 −1 Eb,ϕ1 = ϕ−1 )({u1 } × F1 ) 1 ({u1 } × F1 ) = (ϕ2 ◦ α −1 = ϕ2 ({u2 } × F2 ) = Eb,ϕ2 .

Hence Eb,ϕ1 = Eb,ϕ2 as sets, and it is easily seen that addition and scalar multiplication in Eb,ϕ1 and Eb,ϕ2 are identical as are the topologies. For (ii) note that if v ∈ E, −1 ϕ1 (v) = (u1 , f1 ), ϕ2 (v) = (u2 , f2 ), b1 = ϕ−1 1 (u1 , 0), and b2 = ϕ2 (u2 , 0),

then ψ21 (u2 , f2 ) = (u1 , f1 ), so ψ21 gives a linear isomorphism {u2 } × F2 → {u1 } × F1 , and therefore ϕ1 (b2 ) = ψ21 (u2 , 0) = (u1 , 0) = ϕ1 (b1 ), or b2 = b1 . To prove (iii) we verify that for b ∈ B there is an admissible chart with the submanifold property. To get such a manifold chart, we choose an admissible vector bundle chart (W, ϕ), b ∈ W . Then ϕ(W ∩ B) = U × {0} = ϕ(W ) ∩ (E × {0}). Finally, for (iv), it is enough to check that π is C ∞ using local bundle charts. But this is clear, for such a representative is of the form (u, f ) 7→ (u, 0). That π is onto is clear.  The fibers of a vector bundle inherit an intrinsic vector space structure and a topology independent of the charts, but there is no norm that is chart independent. Putting particular norms on fibers is extra structure to be considered later in the book. Sometimes the phrase Banachable space is used to indicate that the topology comes from a complete norm but we are not nailing down a particular one. The following summarizes the basic properties of a vector bundle.

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3. Manifolds and Vector Bundles

3.4.7 Theorem. Let E be a vector bundle. The zero section (or base) B of E is a submanifold of E and there is a map π : E → B (sometimes denoted πBE : E → B) called the projection that is of class C ∞ , and is surjective (onto). Moreover, for each b ∈ B, π −1 (b), called the fiber over b, has a Banachable vector space structure induced by any admissible vector bundle chart, with b the zero element. Because of these properties we sometimes write “the vector bundle π : E → B” instead of “the vector bundle (E, V).” Fibers are often denoted by Eb = π −1 (b). If the base B and the map π are understood, we just say “the vector bundle E.” Tangent Bundle as a Vector Bundle. A commonly encountered vector bundle is the tangent bundle τM : T M → M of a manifold M . To see that the tangent bundle, as we defined it in the previous section, is a vector bundle in the sense of this section, we use the following lemma. 3.4.8 Lemma. If f : U ⊂ E → V ⊂ F is a diffeomorphism of open sets in Banach spaces, then T f : U × E → V × F is a local vector bundle isomorphism. Proof. Since T f (u, e) = (f (u), Df (u) · e), T f is a local vector bundle mapping. But as f is a diffeomorphism, (T f )−1 = T (f −1 ) is also a local vector bundle mapping, and hence T f is a vector bundle isomorphism.  Let A = {(U, ϕ)} be an atlas of admissible charts on a manifold M that is modeled on a Banach space E. In the previous section we constructed the atlas T A = {(T U, T ϕ)} of the manifold T M . If Ui ∩ Uj 6= ∅, then the overlap map −1 T ϕi ◦ T ϕ−1 j = T (ϕi ◦ ϕj ) : ϕj (Ui ∩ Uj ) × E → ϕi (Ui ∩ Uj ) × E

has the expression −1 (u, e) 7→ ((ϕi ◦ ϕ−1 j )(u), D(ϕi ◦ ϕj )(u) · e).

By Lemma 3.4.8, T (ϕi ◦ ϕ−1 j ) is a local vector bundle isomorphism. This proves the first part of the following theorem. 3.4.9 Theorem.

Let M be a manifold and A = {(U, ϕ)} be an atlas of admissible charts.

(i) Then T A = {(T U, T ϕ)} is a vector bundle atlas of T M , called the natural atlas. −1 (ii) If m ∈ M , then τM (m) = Tm M is a fiber of T M and its base B is diffeomorphic to M by the map τM |B : B → M .

Proof. (ii) Let (U, ϕ) be a local chart at m ∈ M , with ϕ : U → ϕ(U ) ⊂ E and ϕ(m) = u. Then T ϕ : T M |U → ϕ(u) × E is a natural chart of T M , so that T ϕ−1 ({u} × E) = T ϕ−1 { [cu,e ]u | e ∈ E } by definition of T ϕ, and this is exactly Tm M . For the second assertion, τM |B is obviously a bijection, and its local representative with respect to T ϕ and ϕ is the natural identification determined by ϕ(U )×{0} → ϕ(U ), a diffeomorphism.  Thus, Tm M is isomorphic to the Banach space E, the model space of M , M is identified with the zero section of T M , and τM is identified with the bundle projection onto the zero section. It is also worth recalling that the local representative of τM is (ϕ ◦ τM ◦ T ϕ−1 )(u, e) = u, that is, just the projection of ϕ(U ) × E to ϕ(U ). 3.4.10 Examples. A.

Any manifold M is a vector bundle with zero-dimensional fiber, namely M × {0}.

3.4 Vector Bundles

157

b

Figure 3.4.3. The cylinder as a vector bundle

B. The cylinder E = S 1 × R is a vector bundle with π : E → B = S 1 the projection on the first factor (Figure 3.4.3). This is a trivial vector bundle in the sense that it is a product. The cylinder is diffeomorphic to T S 1 by Example 3.3.14A. C. The M¨ obius band is a vector bundle π : M → S 1 with one-dimensional fiber obtained in the following way (see Figure 3.4.4). On the product manifold R × R, consider the equivalence relation defined by (u, v) ∼ (u0 , v 0 ) iff u0 = u+k, v 0 = (−1)k v for some k ∈ Z and denote by p : R×R → M the quotient topological space. Since the graph of this relation is closed and p is an open map, M is a Hausdorff space. Let [u, v] = p(u, v) and define the projection π : M → S 1 by π[u, v] = e2πiu . Let V1 = ]0, 1[ × R, V2 = ](−1/2), (1/2)[ × R, U1 = S 1 \{1}, and U2 = S 1 \{−1} and then note that p|V1 : V1 → π −1 (U1 ) and p|V2 : V2 → π −1 (U2 ) are homeomorphisms and that M = π −1 (U1 ) ∪ π −1 (U2 ). Let {(U1 , ϕ1 ), (U2 , ϕ2 )} be an atlas with two charts for S 1 (see Example 3.1.2). Define ψj : π −1 (Uj ) → R × R

by ψj = χj ◦ (p|Vj )−1

and χj : Vj → R × R

by χj (u, v) = (ϕj (e2πiu ), (−1)j+1 v), j = 1, 2

and observe that χj and ψj are homeomorphisms. Since the composition ψ2 ◦ ψ1−1 : (R × R)\({0} × R) → (R × R)\({0} × R) is given by the formula (ψ2 ◦ ψ1−1 )(x, y) = ((ϕ2 ◦ ϕ−1 1 )(x), −y), we see that {(π −1 (U1 ), ψ1 ), (π −1 (U2 ), ψ2 )} forms a vector bundle atlas of M. D. The Grassmann bundles (universal bundles). γn (E) → Gn (E),

We now define vector bundles

γ n (E) → Gn (E),

and γ(E) → G(E),

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3. Manifolds and Vector Bundles

Figure 3.4.4. The M¨ obius band

which play an important role in the classification of isomorphism classes of vector bundles (see for example Hirsch [1976]). The definition of the projection ρ : γn (E) → Gn (E) is the following (see Example 3.1.8G for notations): we let γn (E) = { (F, v) | F is an n-dimensional subspace of E and v ∈ F }, we set ρ(F, v) = F. Our claim is that this defines a vector bundle over Gn (E). The intuition for the case of lines in R3 is very simple: here γ1 (R3 ) is just the space of pairs (`, x), where ` is a line through the origin and x is a point on `. That is, roughly speaking, γ1 (R3 ) is the set of “marked lines” in R3 ; the map ρ is just the map that sends the marked lines into unmarked lines regarded as points in G1 (R3 ), or what is the same thing, RP3 . Note that the fiber of this map ρ over a line ` is the set of marks on that line, which is a copy of the real line R3 . Now we turn to the technical proof of the vector bundle structure. We claim that the charts (ρ−1 (UG ), ψFG ), where E = F ⊕ G, ψFG (H, v) = (ϕFG (H), πG (H, F)(v)), and ψFG : ρ−1 (UG ) → L(F, G) × F, define a vector bundle structure on γn (E). This is because the overlap maps are
−1 ψF0 G0 ◦ ψFG (T, f ) = (ϕF0 G0 ◦ ϕ−1 FG )(T ) ,
(πG0 (graph(T ), F0 ) ◦ πG (graph(T ), F)−1 )(f ) , where T ∈ L(F, G), f ∈ F, and graph(T ) denotes the graph of T in E × F; smoothness in T is shown as in Example 3.1.8G. The fiber dimension of this bundle is n. A similar construction holds for Gn (E) yielding γ n (E); the fiber codimension in this case is also n. Similarly γ(E) → G(E) is obtained with not necessarily isomorphic fibers at different points of G(E).  Vector Bundle Maps.

Now we are ready to look at maps between vector bundles.

3.4.11 Definition. Let E and E 0 be two vector bundles. A map f : E → E 0 is called a C r vector bundle mapping (local isomorphism) when for each v ∈ E and each admissible local bundle chart (V, ψ) of E 0 for which f (v) ∈ V , there is an admissible local bundle chart (W, ϕ) with f (W ) ⊂ W 0 such that the local representative fϕψ = ψ ◦ f ◦ ϕ−1 is a C r local vector bundle mapping (local isomorphism). A bijective local vector bundle isomorphism is called a vector bundle isomorphism. This definition makes sense only for local vector bundle charts and not for all manifold charts. Also, such a W is not guaranteed by the continuity of f , nor does it imply it. However, if we first check that f is fiber preserving (which it must be) and is continuous, then such an open set W is guaranteed. This fiber–preserving character is made more explicit in the following. 3.4.12 Proposition.

Suppose f : E → E 0 is a C r vector bundle map, r ≥ 0. Then:

(i) f preserves the zero section: f (B) ⊂ B 0 ; (ii) f induces a unique mapping fB : B → B 0 such that the following diagram commutes:

3.4 Vector Bundles

E

f

159

- E0 π0 ? - B0

π ? B fB

that is, π 0 ◦ f = fB ◦ π. (Here, π and π 0 are the projection maps.) Such a map f is called a vector bundle map over fB . (iii) A C ∞ map g : E → E 0 is a vector bundle map iff there is a C ∞ map gB : B → B 0 such that π 0 ◦ g = gB ◦ π and g restricted to each fiber is a linear continuous map into a fiber. Proof. (i) Suppose b ∈ B. We must show f (b) ∈ B 0 . That is, for a vector bundle chart (V, ψ) with f (b) ∈ V we must show ψ(f (b)) = (v, 0). Since we have a chart (W, ϕ) such that b ∈ W , f (W ) ⊂ V , and ϕ(b) = (u, 0), it follows that ψ(f (b)) = (ψ ◦ f ◦ ϕ−1 )(u, 0) which is of the form (v, 0) by linearity of fϕψ on each fiber. For (ii), let fB = f |B : B → B 0 . With the notations above, 0 ψ|B 0 ◦ π 0 ◦ f ◦ ϕ−1 = πψ,ψ|B 0 ◦ fϕψ

and ψ|B 0 ◦ fB ◦ π ◦ ϕ−1 = (fB )ϕ|B,ψ|B 0 ◦ πϕ,ϕ|B which are equal by (i) and because the local representatives of π and π 0 are projections onto the first factor. Also, if fϕψ = (α1 , α2 ), then (fB )ϕψ = α1 , so fB is C r . One half of (iii) is clear from (i) and (ii). For the converse we see that in local representation, g has the form gϕψ (u, f ) = (ψ ◦ g ◦ ϕ−1 )(u, f ) = (α1 (u), α2 (u) · f ), which defines α1 and α2 . Since g is linear on fibers, α2 (u) is linear. Thus, the local representatives of g with respect to admissible local bundle charts are local bundle mappings by Proposition 3.4.3.  We also note that the composition of two vector bundle mappings is again a vector bundle mapping. 3.4.13 Examples. A. Let M and N be C r+1 manifolds and f : M → N a C r+1 map. Then T f : T M → T N is a C r vector bundle map of class C r . Indeed the local representative of T f , (T f )T ϕ,T ψ = T (fϕψ ) is a local vector bundle map, so the result follows from Proposition 3.3.11. B. The proof of Proposition 3.3.13 shows that T (M1 × M2 ) and T M1 × T M2 are isomorphic as vector bundles over the identity of M1 × M2 . They are usually identified. C. To get an impression of how vector bundle maps work, let us show that the cylinder S 1 × R and the M¨obius band M are not vector bundle isomorphic. If ϕ : M → S 1 × R were such an isomorphism, then the image of the curve c : [0, 1] → M, c(t) = [t, 1] by ϕ would never cross the zero section in S 1 × R, since [s, 1] is never zero in all fibers of M; that is, the second component of (ϕ ◦ c)(t) 6= 0 for all t ∈ [0, 1]. But c(1) = [1, 1] = [0, −1] = −[0, 1] = −c(0) so that the second components of ϕ ◦ c at t = 0 and t = 1 are in absolute value equal and of opposite sign, which, by the intermediate value theorem, implies that the second component of ϕ ◦ c vanishes somewhere.

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3. Manifolds and Vector Bundles

D. It is shown in differential topology that for any vector bundle E with an n-dimensional base B and k-dimensional fiber there exists a vector bundle map ϕ : E → B × Rp , where p ≥ k + n, with ϕB = IB and which, when restricted to each fiber, is injective (Hirsch [1976]). Write ϕ(v) = (π(v), F (v)) so F : E → Rp is linear on fibers. With the aid of this theorem, analogous in spirit to the Whitney embedding theorem, we can construct a vector bundle map Φ : E → γk (Rp ) by Φ(v) = (F (Eb ), F (v)) where v ∈ Eb . Note that ΦB : B → Gk (Rp ) maps b ∈ B to the k-plane F (Eb ) in Rp . Furthermore, note that E is vector bundle isomorphic to the pull-back bundles Φ∗ (γk (Rp )) (see Exercise 3.4-15 for the definition of pull–back bundles). It is easy to check that ϕ 7→ Φ is a bijection. Mappings f : B → Gk (Rp ) such that f ∗ (γk (Rp )) is isomorphic to E are called classifying maps for E; they play a central role in differential topology since they convert the study of vector bundles to homotopy theory (see Hirsch [1976] and Husemoller [1966]).  Sections of Vector Bundles. A second generalization of a local C r mapping, f : U ⊂ E → F, globalizes not f but rather its graph mapping λf : U → U × F; u 7→ (u, f (u)). 3.4.14 Definition. Let π : E → B be a vector bundle. A C r local section of π is a C r map ξ : U → E, where U is open in B, such that for each b ∈ U , π(ξ(b)) = b. If U = B, ξ is called a C r global section, or simply a C r section of π. Let Γr (π) denote the set of all C r sections of π, together with the obvious real (infinite-dimensional ) vector space structure. The condition on ξ says that ξ(b) lies in the fiber over b. The C r sections form a linear function space suitable for global linear analysis. As will be shown in later chapters, this general construction includes spaces of vector and tensor fields on manifolds. The space of sections of a vector bundle differs from the more general class of global C r maps from one manifold to another, which is a nonlinear function space. (See, for example, Eells [1958], Palais [1968], Elliasson [1967], or Ebin and Marsden [1970] for further details.) Subbundles. bundles.

Submanifolds were defined in the preceding section. There are two analogies for vector

3.4.15 Definition. If π : E → B is a vector bundle and M ⊂ B a submanifold, the restricted bundle πM : EM = E|M → M is defined by [ EM = Em , πM = π|EM . m∈M

The restriction πM : EM → M is a vector bundle whose charts are induced by the charts of E in the following way. Let (V, ψ1 ), ψ1 : V → V 0 ⊂ E0 × {0}, be a chart of M induced by the chart (U, ϕ1 ) of B with the submanifold property, where (π −1 (U ), ϕ) (with ϕ(e) = (ϕ1 (π(e)), ϕ2 (e)), ϕ : π −1 (U ) → U 0 × F, and U 0 ⊂ E0 × E00 = E) is a vector bundle chart of E. Then −1 ψ : πM (V ) → V 0 × F,

ψ(e) = (ψ1 (π(e)), ϕ2 (e))

defines a vector bundle chart of EM . It can be easily verified that the overlap maps satisfy VB2. For example, any vector bundle, when restricted to a vector bundle chart domain of the base, defines a vector bundle that is isomorphic to a local vector bundle. 3.4.16 Definition. Let π : E → B be a vector bundle. A subset F ⊂ E is called a subbundle if for each b ∈ B there is a vector bundle chart (π −1 (U ), ϕ) of E where b ∈ U ⊂ B and ϕ : π −1 (U ) → U 0 × F, and a split subspace G of F such that ϕ(π −1 (U ) ∩ F) = U 0 × (G × {0}). These induced charts are verified to form a vector bundle atlas for π|F : F → B. Note that subbundles have the same base as the original vector bundle. Intuitively, the restriction cuts the base keeping the fibers intact, while a subbundle has the same base but smaller fiber, namely Fb = F ∩ Eb . Note that a subbundle F is a closed submanifold of E. For example γk (Rn ) is a subbundle of both γk (Rn+1 ) and γk+1 (Rn+1 ), the canonical inclusions being given by (F, x) 7→ (F × {0}, (x, 0)) and (F, x) 7→ (F × R, (x, 0)), respectively.

3.4 Vector Bundles

Quotients, Kernels, and Ranges. dles and maps.

161

We now consider some additional basic operations with vector bun-

3.4.17 Proposition. Let π : E → B be a vector bundle and F ⊂ E a subbundle. Consider the following equivalence relation on E : v ∼ v 0 if there is a b ∈ B such that v, v 0 ∈ Eb and v − v 0 ∈ Fb . The quotient set E/∼ has a unique vector bundle structure for which the canonical projection p : E → E/∼ is a vector bundle map over the identity. This vector bundle is called the quotient E/F and has fibers (E/F )b = Eb /Fb . Proof. Since F ⊂ E is a subbundle there is a vector bundle chart ϕ : π −1 (U ) → U 0 ×F and split subspaces F1 , F2 , F1 ⊕ F2 = F, such that ϕ|π −1 (U ) ∩ ∗F : (π|F )−1 (U ) → U 0 × (F1 × {0}) is a vector bundle chart for F . The map π induces a unique map Π : E/∼ → B such that Π ◦ p = π. Similarly ϕ induces a unique map Φ : Π−1 (U ) → U 0 × ({0} × F2 ) by the condition Φ ◦ p = ϕ|ϕ−1 (U 0 × ({0} × F2 )), which is seen to be a homeomorphism. One verifies that the overlap map of two such Φ is a local vector bundle isomorphism, thus giving a vector bundle structure to E/∼, with fiber Eb /Fb , for which p : E → E/∼ is a vector bundle map. From the definition of Φ it follows that the structure is unique if p is to be a vector bundle map over the identity.  3.4.18 Proposition. Let π : E → B and ρ : F → B be vector bundles over the same manifold B and f : E → F a vector bundle map over the identity. Let fb : Eb → Fb be the restriction of f to the fiber over b ∈ B and define the kernel of f by [ ker(f ) = ker(fb ) b∈B

and the range of f by range(f ) =

[

range(fb ).

b∈B

(i) ker(f ) and range(f ) are subbundles of E and F respectively iff for every b ∈ B there are vector bundle charts (π −1 (U ), ϕ) of E and (ρ−1 (U ), ψ) of F such that the local representative of f has the form fϕψ : U 0 × (F1 × F2 ) → U 0 × (G1 × G2 ), where fϕψ (u, (f1 , f2 )) = (u, (χ(u) · f2 , 0)), and χ(u) : F2 → G1 is a continuous linear isomorphism. (ii) If E has finite-dimensional fiber, the condition in (i) is equivalent to the local constancy of the rank of the linear map fb : Eb → Fb . Proof. (i) It is enough to prove the result for local vector bundles. But there it is trivial since ker(fϕψ )u = F1 and range(fϕψ )u = G1 . (ii) Fix u ∈ U 0 and put (fϕψ )u (F) = G1 . Then since G1 is closed and finite dimensional in G, it splits; let G = G1 ⊕ G2 . Let F1 = ker(fϕψ )u ; F is finite dimensional and hence F = F1 ⊕ F2 . Then (fϕψ )u : F2 → G1 is an isomorphism. Write  0      a(u ) b(u0 ) F1 G1 (fϕψ )u0 = : → c(u0 ) d(u0 ) F2 G2 for u0 ∈ U 0 and note that b(u0 ) is an isomorphism. Therefore b(u0 ) is an isomorphism for all u0 in a neighborhood of u by Lemma 2.5.4. We can assume that this neighborhood is U 0 , by shrinking U 0 if necessary. Note also that a(u) = 0, c(u) = 0, d(u) = 0. The rank of (fϕψ )u0 is constant in a neighborhood of u, so

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shrink U 0 further, if necessary, so that (fϕψ )u0 has constant rank for all u0 ∈ U 0 . Since b(u0 ) is an isomorphism, a(u0 )(F1 ) + b(u0 )(F2 ) = G1 and since the rank of (fϕψ )u0 equals the dimension of G1 , it follows that c(u0 ) = 0 and d(u0 ) = 0 for all u0 ∈ U 0 . Then   I 0 ∈ GL(F1 ⊕ F2 , F1 ⊕ F2 ) λ u0 = −b(u0 )−1 a(u0 ) I and (fϕψ )u0 ◦ λu0 =

  0 b(u0 ) 0 0

which yields the form of the local representative in (i) after fiberwise composing ϕu0 with λ−1 u0 . f



g

3.4.19 Definition. A sequence of vector bundle maps over the identity E → F → G is called exact at F if range(f ) = ker(g). It is split fiber exact if ker(f ), range(g), and range(f ) = ker(g) split in each fiber. It is bundle exact if it is split fiber exact and ker(f ), range(g), and range(f ) = ker(g) are subbundles. 3.4.20 Proposition.

Let E, F , and G be vector bundles over a manifold B and let f

g

E −→ F −→ G be a split fiber exact sequence of smooth bundle maps. Then the sequence is bundle exact; that is, ker(f ), range(f ) = ker(g), and range(g) are subbundles of E, F , and G respectively. Proof. Fixing b ∈ B, set A = ker(fb ), B = ker(gb ) = range(fb ), C = range(gb ), and let D be a complement for C in Gb , so Eb = A × B, Fb = B × C, and Gb = C × D. Let ϕ : U → U 0 be a chart on B at b, ϕ(b) = 0, defining vector bundle charts on E, F , and G. Then the local representatives f 0 : U 0 × A × B → U 0 × B × C,

g0 : U 0 × B × C → U 0 × C × D

of f and g respectively are the identity mappings on U 0 and can be written as matrices of operators       z w A B 0 fu0 = : → x y B C and gu0 0



β = α

     γ B C : → δ C D

depending smoothly on u0 ∈ U 0 . Now since w0 and γ0 are isomorphisms by Banach’s isomorphism theorem, shrink U and U 0 such that wu0 and γu0 are isomorphisms for all u0 ∈ U 0 . By exactness, gu0 0 ◦ fu0 0 = 0, which in terms of the matrix representations becomes x = −γ −1 ◦ β ◦ z,

y = −γ −1 ◦ β ◦ w,

α = −δ ◦ y ◦ w−1 ,

that is, x = y ◦ w−1 ◦ z

and

α = δ ◦ γ −1 ◦ β.

Extend fu0 0 to the map hu0 : A × B × C → A × B × C depending smoothly on u0 ∈ U 0 by   I 0 0 hu 0  z w 0  . x y I We find maps a, b, c, d, k, m, n, p such that    I 0 0 k 0 a b  hu0 m 0 c d 0

p n 0

 0 0 = I, I

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163

which can be accomplished by choosing k = I,

d = I,

p = 0,

n = I,

m = −w−1 ◦ z,

b = 0,

a = w−1 ,

c = −y ◦ w−1 ,

and taking into account that x = y ◦ w−1 ◦ z. This procedure gives isomorphisms         a b w−1 0 B B λ= = : → c d −y ◦ w−1 I C C         k p I 0 A A µ= = : → m n −w−1 ◦ z I B B depending smoothly on u0 ∈ U such that λ ◦ fu0 0 ◦ µ =

      0 I A B : → . 0 0 B C

Proposition 3.4.18(ii) shows that ker(f ) and range(f ) are subbundles. The same procedure applied to gu0 0 proves that ker(g) and range(g) are subbundles and thus the fiber split exact sequence f

g

E→F →G is bundle exact.

 g

As a special case note that 0 → F → G is split fiber exact when gb is injective and has split range. Here 0 is the trivial bundle over B with zero-dimensional fiber and the first arrow is injection to the zero f section. Similarly, taking G = 0 and g the zero map, the sequence E → F → 0 is split fiber exact when fb is surjective with split kernel. In both cases range(g) and ker(f ) are subbundles by Proposition 3.4.20. In Proposition 3.4.20, and these cases in particular, we note that if the sequences are split fiber exact at b, then they are also split fiber exact in a neighborhood of b by the openness of GL(E, E) in L(E, E). A split fiber exact sequence of the form f

g

0 −→ E −→ F → G → 0 is called a short exact sequence. By Proposition 3.4.20 and Proposition 3.4.17, any split fiber exact sequence f

g

E −→ F −→ G induces a short exact sequence [f ]

g

0 −→ E/ ker(f ) −→ F → range(g) → 0 where [f ]([e]) = f (e) for e ∈ E. 3.4.21 Definition.

A short exact sequence f

g

0 −→ E −→ F −→ G −→ 0 h

is said to be split exact if there is a split fiber exact sequence 0 → G → F such that g ◦ h is the identity on G.

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Products and Tensorial Constructions. The geometric meaning of this concept will become clear after we introduce a few additional constructions with vector bundles. 3.4.22 Definition. If π : E → B and π 0 : E 0 → B 0 are two vector bundles, the product bundle π ×π 0 : E ×E 0 → B ×B 0 is defined by the vector bundle atlas consisting of the sets π −1 (U )×π 0−1 (U 0 ) and the maps ϕ × ψ, where (π −1 (U ), ϕ), U ⊂ B, and (π 0−1 (U 0 ), ψ), U 0 ⊂ B 0 are vector bundle charts of E and E 0 , respectively. It is straightforward to check that the product atlas verifies conditions VB1 and VB2 of Definition 3.4.4. Below we present a general construction, special cases of which are used repeatedly in the rest of the book. It allows the transfer of vector space constructions into vector bundle constructions. The abstract procedure will become natural in the context of examples given below in 3.4.25 and later in the book. 3.4.23 Definition. Let I and J be finite sets and consider two families E = (Ek )k∈I∪J , and E 0 = (E0k )k∈I∪J of Banachable spaces. Let Y Y L(E, E 0 ) = L(Ei , E0i ) × L(E0j , Ej ) i∈I

j∈J

and let (Ak ) ∈ L(E, E 0 ); that is, Ai ∈ L(Ei , E0i ), i ∈ I, and Aj ∈ L(E0j , Ej ), j ∈ J. An assignment Ω taking any family E to a Banach space ΩE and any sequence of linear maps (Ak ) to a linear continuous map Ω(Ak ) ∈ L(ΩE, ΩE 0 ) satisfying Ω(IEk ) = IΩE ,

Ω((Bk ) ◦ (Ak )) = Ω((Bk )) ◦ Ω((Ak ))

(composition is taken componentwise) and is such that the induced map Ω : L(E, E 0 ) → L(ΩE, ΩE 0 ) is C ∞ , will be called a tensorial construction of type (I, J). 3.4.24 Proposition. Let Ω be a tensorial construction of type (I, J) and E = (E k )k∈I∪J be a family of vector bundles with the same base B. Let [ ΩE = ΩEb , where Eb = (Ebk )k∈I∪J . b∈B

Then ΩE has a unique vector bundle structure over B with (ΩE)b = ΩEb and π : ΩE → B sending ΩEb to b ∈ B, whose atlas is given by the charts (π −1 (U ), ψ), where ψ : π −1 (U ) → U 0 × Ω((Fk )) is defined as follows. Let (πk−1 (U ), ϕk ), ϕk : πk−1 (U ) → U 0 × Fk , ϕk (ek ) = (ϕ1 (πk (ek )), ϕk2 (ek )) be vector bundle charts on E k inducing the same manifold chart on B. Define ψ(x) = (ϕ1 (π(x)), Ω(ψπ(x) )(x))

by

k ψπ(x) = (ψπ(x) ),

 −1 j i for j ∈ J. where ψπ(x) = ϕi2 for i ∈ I and ψπ(x) = ϕj2 Proof.

We need to show that the overlap maps are local vector bundle isomorphisms. We have 0k k −1 (ψ 0 ◦ ψ −1 )(u, e) = ((ϕ01 ◦ ϕ−1 (u)) · e)), 1 )(u), Ω((ϕ2 ◦ (ϕ2 )

the first component of which is trivially C ∞ . The second component is also C ∞ since each ϕk is a vector bundle chart by the composite mapping theorem, and by the fact that Ω is smooth.  3.4.25 Examples.

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165

A. Whitney sum. Choose for the tensorial construction the following: J = ∅, I = {1, . . . , n}, and ΩE is the single Banach space E1 × · · · × En . Let Ω((Ai )) = A1 × · · · × An . The resulting vector bundle is denoted by E1 ⊕ · · · ⊕ En and is called the Whitney sum. The fiber over b ∈ B is just the sum of the component fibers. B. Vector bundles of bundle maps. Let E1 , E2 be two vector bundles. Choose for the tensorial construction the following: I, J are one–point sets I = {1}, J = {2}, Ω(A1 , A2 ) · S = A1 ◦ S ◦ A2

Ω(E1 , E2 ) = L(E2 , E1 ),

for S ∈ L(E1 , E1 ). The resulting bundle is denoted by L(E2 , E1 ). The fiber over b ∈ B consists of the linear maps of (E2 )b to (E1 )b . C. Dual bundle. This is a particular case of Example B for which E = E2 and E1 = B ×R. The resulting bundle is denoted E ∗ ; the fiber over b ∈ B is the dual Eb∗ . If E = T M , then E ∗ is called the cotangent bundle of M and is denoted by T ∗ M . D. Vector bundle of multilinear maps. Let E0 , E1 , . . . , En be vector bundles over the same base. The space of n-multilinear maps (in each fiber) L(E1 , . . . , En ; E0 ) is a vector bundle over B by the choice of the following tensorial construction: I = {0}, J = {1, . . . , n}, Ω(E0 , . . . , En ) = Ln (E1 , . . . , En ; E0 ), Ω(A0 , A1 , . . . , An ) · S = A0 ◦ S ◦ (A1 × · · · × An ) n

for S ∈ L (E1 , . . . , En ; E0 ). One may similarly construct Lks (E; E0 ) and Lka (E; E0 ), the vector bundle of symmetric and antisymmetric k-linear vector bundle maps of E × E × · · · × E to E0 .  3.4.26 Proposition.

A short exact sequence of vector bundles f

g

0 −→ E −→ F → G → 0 is split if and only if there is a vector bundle isomorphism ϕ : F → E ⊕ G such that ϕ ◦ f = i and p ◦ ϕ = g, where i : E → E ⊕ G is the inclusion u 7→ (u, 0) and p : E ⊕ G → G is the projection (u, w) 7→ w. Proof.

Note that i

p

0→E →E⊕G→G→0 is a split exact sequence; the splitting is given by w ∈ G 7→ (0, w) ∈ E ⊕ G. If there is an isomorphism ϕ : F → E ⊕ G as in the statement of the proposition, define h : G → F by h(w) = ϕ−1 (0, w). Since ϕ is an h

isomorphism and G is a subbundle of E ⊕ G, it follows that 0 → G → F is split fiber exact. Moreover (g ◦ h)(w) = (g ◦ ϕ−1 )(0, w) = p(0, w) = w. Conversely, assume that h

0 −→ G −→ F

is a splitting of

f

g

0 −→ E −→ F −→ G −→ 0,

that is, g ◦ h = identity on G. Then range(h) is a subbundle of F (by Definition 3.4.19) which is isomorphic to G by h. Since g ◦ h = identity, it follows that range(h) ∩ ker(g) = 0. Moreover, since any v ∈ F can be written in the form v = (v − h(g(v))) + h(g(v)), with h(g(v)) ∈ range(h) and v − h(g(v)) ∈ ker(g), it follows that F = ker(g) ⊕ range(h). Since the inverse of ϕ is given by (u, v) 7→ (f (u), h(v)), it follows that the map ϕ is a smooth vector bundle isomorphism and that the identities ϕ ◦ f = i, p ◦ ϕ = g hold. 

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3. Manifolds and Vector Bundles

Fiber Bundles. We next give a brief account of a useful generalization of vector bundles, the locally trivial fiber bundles. 3.4.27 Definition. A C k fiber bundle, where k ≥ 0, with typical fiber F (a given manifold ) is a C k surjective map of C k manifolds π : E → B which is locally a product, that is, the C k manifold B has an open atlas {(Uα , ϕα )}α∈A such that for each α ∈ A there is a C k diffeomorphism χα : π −1 (Uα ) → Uα × F such that pα ◦ χα = π, where pα : Uα × F → Uα is the projection. The C k manifolds E and B are called the total space and base of the fiber bundle, respectively. For each b ∈ B, π −1 (b) = Eb is called the fiber over b. The C k diffeomorphisms χα are called fiber bundle charts. If k = 0, E, B, F are required to be only topological spaces and {Uα } an open covering of B. Each fiber Eb = π −1 (b), for b ∈ B, is a closed C k submanifold of E, which is C k diffeomorphic to F via χα |Eb . The total space E is the disjoint union of all of its fibers. By the local product property, the C k manifold structure of E is given by an atlas whose charts are products, that is, any chart on E contains a chart of the form ραβ = (ϕα × ψβ ) ◦ χα : χ−1 α (Uα × Vβ ) → ϕα (Uα ) × ψβ (Vβ ), where (Uα , ϕα ) is a chart on B satisfying the property of the definition and thus giving rise to χα , and (Vβ , ψβ ) is any chart on F . Note that the maps χαb = χα |Eb : Eb → F are C k diffeomorphisms. If (Uα0 , ϕα0 ) and χα0 are as in Definition 3.4.27 with Uα ∩ Uα0 6= ∅, then the diffeomorphism χα0 ◦ χ−1 α : (Uα ∩ Uα0 ) × F → (Uα ∩ Uα0 ) × F is given by −1 (χα0 ◦ χ−1 α )(u, f ) = (u, (χα0 u ◦ χαu )(f )) k and therefore χα0 u ◦ χ−1 αu : F → F is a C diffeomorphism. This proves the uniqueness part in the following proposition.

3.4.28 Proposition. Assume that

Let E be a set, B and F be C k manifolds, and let π : E → B be a surjective map.

(i) there is a C k atlas {(Uα , ϕα )} of B and a family of bijective maps χα : π −1 (Uα ) → Uα × F satisfying pα ◦ χα = π, where pα : Uα × F → Uα is the projection, and that k (ii) the maps χα0 ◦ χ−1 α : Uα × F → Uα0 × F are C diffeomorphisms whenever Uα0 ∩ Uα 6= ∅.

Then there is a unique C k manifold structure on E for which π : E → B is a C k locally trivial fiber bundle with typical fiber F . Proof. Define the atlas of E by (χ−1 α (Uα ×Vβ ), ραβ ), where (Uα , ϕα ) is a chart in the atlas of B given in (i), χα : π −1 (Uα ) → Uα ×F is the bijective map given in (i), (Vβ , ψβ ) is any chart on F , and ραβ = (ϕα ×ψβ )◦χα . If (Uα0 , ϕα0 ) is another chart of the atlas of B in (i) and (Vβ 0 , ψβ 0 ) is another chart on F such that Uα ∩Uα0 6= ∅ and Vβ ∩ Vβ 0 6= ∅, then the overlap map −1 −1 −1 0 0 0 ρα0 β 0 ◦ ρ−1 αβ = (ϕα × ψβ ) ◦ χα ◦ χα ◦ (ϕα × ψβ ) k k is C k by (i). Thus {(χ−1 α (Uα ×Vβ ), ραβ )} is a C atlas on E relative to which π : E → B is a C locally trivial fiber bundle by (i). The differentiable structure on E is unique by the remarks preceding this proposition. 

Many of the concepts introduced for vector bundles have generalizations to fiber bundles. For instance, local and global sections are defined as in Definition 3.4.14. Given a fiber bundle π : E → B, the restricted bundle πM : EM = E|M → M , for M a submanifold of B is defined as in Definition 3.4.15. A locally trivial subbundle of π : E → F with typical fiber G, a submanifold of F , is a submanifold E 0 of E such that the map π 0 = π|E 0 : E 0 → B is onto and satisfies the following property: if χα : π −1 (Uα ) → Uα × F is a local trivialization of E, then χ0α = χα |π 0−1 (Uα ) → Uα × G are local trivializations. Thus π 0 : E 0 → B

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167

is a locally trivial fiber bundle in its own right. Finally, locally trivial fiber bundle maps, or fiber bundle morphisms are defined in the following way. If π 0 : E 0 → B 0 is another locally trivial fiber bundle with typical fiber F 0 , then a smooth map f : E → E 0 is called fiber preserving if π(e1 ) = π(e2 ) implies (π 0 ◦f )(e1 ) = (π 0 ◦f )(e2 ), for e1 , e2 ∈ E. Thus f determines a map fB : B → B 0 satisfying π 0 ◦f = π ◦fB . The map fB is smooth since for any chart (Uα , ϕα ) of B inducing a local trivialization χα : π −1 (Uα ) → Uα × F , the map fB can be written as fB (b) = (π ◦ f ◦ χ−1 α )(b, n), for any fixed n ∈ F . The pair (f, fB ) is called a locally trivial fiber bundle map or fiber bundle morphism. An invertible fiber bundle morphism is called a fiber bundle isomorphism. 3.4.29 Examples. A.

Any manifold is a locally trivial fiber bundle with typical fiber a point.

B. Any vector bundle π : E → B is a locally trivial fiber bundle whose typical fiber is the model of the fiber Eb . Indeed, if ϕ : W → U 0 × F, where U 0 open in E, is a local vector bundle chart, by Proposition 3.4.6, ϕ|ϕ−1 (U × {0}) : U → U 0 ⊂ E, U = W ∩ B, is a chart on the base B and χ : π −1 (U ) → U × F defined by χ(e) = (π(e), (p2 ◦ ϕ)(e)), where p2 : U 0 × F → F is the projection, is a local trivialization of E. In fact, any locally trivial fiber bundle π : E → B whose typical fiber F is a Banach space is a vector bundle, iff the maps χαb : Eb → F induced by the local trivializations χα : π −1 (Uα ) → Uα × F, are linear and continuous. Indeed, under these hypotheses, the vector bundle charts are given by (ϕα × idF ) ◦ χα : π −1 (Uα ) → Uα × F, where idF is the identity mapping on F. C. Many of the topological properties of a vector bundle are determined by its fiber bundle structure. For example, a vector bundle π : E → B is trivial if and only if it is trivial as a fiber bundle. Clearly, if E is a trivial vector bundle, then it is also a trivial fiber bundle. The converse is also true, but requires topological ideas beyond the scope of this book. (See, for instance, Steenrod [1957].) D. The Klein bottle K (see Figure 1.4.2) is a locally trivial fiber bundle π : K → S 1 with typical fiber S 1 . The space K is defined as the quotient topological space of R2 by the relation (a, b) ∼ (a + k, (−1)k b + n) for all k, n ∈ Z. Let p : R2 → K be the projection p(a, b) = [a, b] and define the surjective map π : K → S 1 by π([a, b]) = e2πia . Let { (Uj , ϕj ) | j = 1, 2 } be the atlas of S 1 given in Example 3.1.2, that is, ϕj : S 1 \{(0, (−1)j+1 )} → R,

ϕj (x, y) =

y , 1 − (−1)j x

which satisfy (ϕ2 ◦ ϕ−1 1 )(z) = 1/z, for z ∈ R\{0}. Define χj : π −1 (Uj ) → Uj × S 1

by χj ([a, b]) = (e2πia , e2πib )

and note that pj ◦ χj = π, where pj : Uj × S 1 → Uj is the projection. Since 1 1 1 1 χ2 ◦ χ−1 1 : (S \{(0, 1)}) × S → (S \{(0, −1)}) × S

is the identity, Proposition 3.4.28 implies that K is a locally trivial fiber bundle with typical fiber S 1 . Further topological results show that this bundle is nontrivial; see Exercise 3.4-16. (Later we will prove that K is non-orientable—see Chapter 7.) E. Consider the smooth map πn : S n → RPn which associates to each point of S n the line through the origin it determines. Then πn : S n → RPn is a locally trivial fiber bundle whose typical fiber is a two-point set. This is easily seen by taking for each pair of antipodal points two small antipodal disks and projecting them to an open set U in RPn ; thus πn−1 (U ) consists of the disjoint union of these disks and the fiber bundle charts simply send this disjoint union to itself. This bundle is not trivial since S n is connected and two disjoint copies of RPn are disconnected. These fiber bundles are also called the real Hopf fibrations.

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3. Manifolds and Vector Bundles

F. This example introduces the classical Hopf fibration h : S 3 → S 2 which is the fibration with the lowest dimensional total space and base among the series of complex Hopf fibrations κn : S 2n+1 → CPn with typical fiber S 1 (see Exercise 3.4-21). To describe h : S 3 → S 2 it is convenient to introduce the division algebra of quaternions H. For x ∈ R4 write x = (x0 , x) ∈ R × R3 and introduce the product (x0 , x)(y 0 , y) = (x0 y 0 − x · y, x0 y + y 0 x + x × y). Relative to this product and the usual vector space structure, R4 becomes a non-commutative field denoted by H and whose elements are called quaternions. The identity element in H is (1, 0), the inverse of (x0 , x) is (x0 , x)−1 = (x0 , −x)/kxk2 , where kxk2 = (x0 )2 + (x1 )2 + (x2 )2 + (x3 )2 . Associativity of the product comes down to the vector identity a × (b × c) = b(a · c) − c(a · b). Alternatively, the quaternions written as linear combinations of the form x0 + ix1 + jx2 + kx3 , where i = (0, i),

j = (0, j),

k = (0, k)

obey the multiplication rules ij = k,

jk = i,

ki = j,

i2 = j2 = k2 = −1.

Quaternions with x0 = 0 are called pure quaternions and the conjugation x 7→ x∗ given by i∗ = −i, j ∗ = −j, k ∗ = −k is an automorphism of the R-algebra H. Then kxk2 = xx∗ and kxyk = kxk kyk for all x, y ∈ H. Finally, the dot product in R4 and the product of H are connected by the relation xz·yz = (x·y)kzk2 , for all x, y, z ∈ H. Fix y ∈ H. The conjugation map cy : H → H defined by cy (x) = yxy −1 is norm preserving and hence orthogonal. Since it leaves the vector (x0 , 0) invariant, it defines an orthogonal transformation of R3 . A simple computation shows that this orthogonal transformation of R3 is given by x 7→ x +

2 [(x · y)y − y 0 (x × y) − (y · y)x] kyk

from which one can verify that its determinant equals one, that is, it is an element of SO(3). Let π : S 3 → SO(3) denote its restriction to the unit sphere in R4 . Choosing x ∈ R3 , define ρx : SO(3) → S 2 by ρx (A) = Ax so that by composition we get hx = ρx ◦ π : S 3 → S 2 . It is easily verified that the inverse image of any point under hx is a circle. Taking for x = −k, minus the third standard basis vector in R3 , hx becomes the standard Hopf fibration h : S 3 → S 2 , h(y 0 , y 1 , y 1 , y 3 ) = (−2y 1 y 3 − 2y 0 y 2 , 2y 0 y 1 − 2y 2 y 3 , (y 1 )2 + (y 2 )2 − (y 0 )2 − (y 3 )2 ) which, by substituting w1 = y 0 + iy 3 , w2 = y 2 + iy 1 ∈ C takes the following classical form H(w1 , w2 ) = (−2w1 w2 , |w2 |2 − |w1 |2 ). Interestingly enough, the Hopf fibration enters into a number of problems in classical mechanics from rigid body dynamics to the dynamics of coupled oscillators (see Marsden and Ratiu [1999], for instance—in fact, the map h above is an example of the important notion of what is called a momentum map)5 . 5 A note for those with an interest in mechanics and momentum maps: regarded as a map C2 → R3 , H is the momentum map (Noether quanitity) for the natural SU (2) action on C2 ; that is, H maps from the space of Cayley-Klein parameters to the rigid body variables. In other communities, the map H is known as the Kustaanheimo-Stiefel (KS) transformation; see Kustaanheimo and Stiefel [1965]. It maps the regular and linear equations of a four-dimensional harmonic oscillator to the singular and nonlinear three-dimensional Kepler problem. The KS transformation is widely used in studies of the perturbed Kepler problem in celestial mechanics and in atomic physics.

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169

G. The Hopf fibration is nontrivial. A rigorous proof of this fact is not so elementary and historically was what led to the introduction of the Hopf invariant, a precursor of the topological theory of StiefelWhitney and Chern classes (Hopf [1931] and Hilton and Wylie [1960]). We shall limit ourselves to a geometric description of this bundle which exhibits its non-triviality.6 In fact we shall describe how each pair of fibers are linked. Cut S 2 along an equator to obtain the closed northern and southern hemispheres, each of which is diffeomorphic to two closed disks DN and DS . Their inverse images in S 3 are two solid tori S 1 × DN and S 1 × DS . We think of S 3 as the compactification of R3 and as the union of two solid tori glued along their common boundary by a diffeomorphism which identifies the parallels of one with meridians of the other and vice-versa. The Hopf fibration on S 3 is then obtained in the following way. Cut each of these two solid tori along a meridian, twist them by 2π and glue each one back together. The result is still two solid tori but whose embedding in R3 is changed: they have the same parallels but twisted meridians; each two meridians are now linked (see Figure 3.4.5). Now glue the two twisted solid tori back together along their common boundary by the diffeomorphism identifying the twisted meridians of one with the parallels of the other and vice-versa, thereby obtaining the total space S 3 of the Hopf fibration. 

Figure 3.4.5. Linked circles in the Hopf fibration

Connectivity Properties of Bundles. Topological properties of the total space E of a locally trivial fiber bundle are to a great extent determined by the topological properties of the base B and the typical fibers F . We present here only some elementary connectivity properties; additional results can be found in Supplement 6.5C and §8.5. 3.4.30 Theorem (Path Lifting Theorem). Let π : E → B be a locally trivial C 0 fiber bundle and let c : [0, 1] → B be a continuous path starting at c(0) = b. Then for each c0 ∈ π −1 (b0 ), there is a unique continuous path c˜ : [0, 1] → E such that c˜(0) = c0 and π ◦ c˜ = c. Proof. Cover the compact set c([0, 1]) by a finite number of open sets Ui , i = 0, 1, . . . , n − 1 such that each χi : π −1 (Ui ) → Ui × F is a fiber bundle chart. Let 0 = t0 < t1 < · · · < tn = 1 be a partition of [0, 1] such that c([ti , ti+1 ]) ⊂ Ui , i = 0, . . . , n − 1. Let χ0 (e0 ) = (b0 , f0 ) and define c0 (t) = χ−1 0 (c(t), f0 )

for t ∈ [0, t0 ].

Then c˜0 is continuous and π ◦ c˜0 = c|[0, t0 ]. Let χ1 (˜ c0 (t1 )) = (c(t1 ), f1 ) and define c˜1 (t) = χ−1 1 (c(t), f1 )

for t ∈ [t1 , t2 ].

6 The Hopf referred to here for the Hopf fibration is “Heinz Hopf”, the brother of the mechanician, “Eberhard Hopf”, famous for the Hopf bifurcation in dynamical systems and fluid mechanics, discovered in the late 1930’s. It is curious that the Hopf fibration has significant roots in mechanics (see the next footnote) and yet, Heinz failed to make any connection to mechanics. This was left to the work of others, notably that of E. Stiefel; see the preceding footnote.

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Then c˜1 is continuous and π ◦ c˜1 = c|[t1 , t2 ]. In addition, if e1 = χ−1 0 (c(t1 ), f0 ) then lim c˜0 (t) = e1

t↑t1

and

lim c˜1 (t) = χ−1 ˜0 (t1 ) = e1 , 1 (c(t1 ), f1 ) = c

t↓t1

that is, the map [0, t2 ] → E which equals c˜0 on [0, t1 ] and c˜1 on [t1 , t2 ] is continuous. Now proceed similarly on U2 , . . . , Un−1 .  Note that if π : E → B is a C k locally trivial fiber bundle and c : [0, 1] → B is a piecewise C k -map, the above construction yields a C k piecewise lift c˜ : [0, 1] → E. 3.4.31 Corollary. Let π : E → B be a C k locally trivial fiber bundle k ≥ 0, with base B and typical fiber F pathwise connected. Then E is pathwise connected. If k ≥ 1, only connectivity of B and F must be assumed. Proof. Let e0 , e1 ∈ E, b0 = π(e0 ), b1 = π(e1 ) ∈ B. Since B is pathwise connected, there is a continuous path c : [0, 1] → B, c(0) = b0 , c(1) = b1 . By Theorem 3.4.30, there is a continuous path c˜ : [0, 1] → E with c˜(0) = e0 . Let c˜(1) = e01 . Since the fiber π −1 (b1 ) is connected there is a continuous path d : [1, 2] → π −1 (b1 ) with d(1) = e01 and d(2) = e1 . Thus γ defined by γ(t) = c˜(t), if t ∈ [0, 1]

and γ(t) = d(t), if t ∈ [1, 2],

is a continuous path with γ(0) = e0 , γ(2) = e1 . Thus E is pathwise connected.



In Supplement 5.5C we shall prove that if π : E → B is a C 0 locally trivial fiber bundle over a paracompact simply connected base with simply connected typical fiber F , then E is simply connected.

Supplement 3.4B Fiber Bundles over Contractible Spaces This supplement proves that any C 0 fiber bundle π : E → B over a contractible base B is trivial. 3.4.32 Lemma. Let π : E → B × [0, 1] be a C 0 fiber bundle. If { Vi | i = 1, . . . , n } is a finite cover of [0, 1] by open intervals such that E|B × Vi is a trivial C 0 fiber bundle, then E is trivial. Proof. By induction it suffices to prove the result for n = 2, that is, prove that if E|B×[0, t] and E|B×[t, 1] are trivial, then E is trivial. If F denotes the typical fiber of E, by hypothesis there are C 0 trivializations over the identity ϕ1 : E|B × [0, t] → B × [0, t] × F and ϕ2 : E|B × [t, 1] → B × [t, 1] × F . The map ϕ2 ◦ ϕ−1 1 : B × {t} × F → B × {t} × F is a homeomorphism of the form (b, t, f ) 7→ (b, t, αb (f )), where αb : F → F is a homeomorphism depending continuously on b. Define the homeomorphism χ : (b, s, f ) ∈ B × [t, 1] × F 7→ (b, s, αb−1 (f )) ∈ B × [t, 1] × F . Then the trivialization χ ◦ ϕ2 : E|B × [t, 1] → B × [t, 1] × F sends any e ∈ π −1 (B × {t}) to ϕ1 (e). Therefore, the map that sends e to the element of B × [0, 1] × F given by ϕ1 (e), if π(e) ∈ B × [0, t] and (χ ◦ ϕ2 )(e), if π(e) ∈ B × [t, 1] is a continuous trivialization of E.  3.4.33 Lemma. Let π : E → B × [0, 1] be a C 0 fiber bundle. Then there is an open covering {Ui } of B such that E|Ui × [0, 1] is trivial.

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171

Proof. There is a covering of B × [0, 1] by sets of the form W × V where W is open in B and V is open in [0, 1], such that E|W × V is trivial. For each b ∈ B consider the family Φb of sets W × V for which b ∈ W . By compactness of [0, 1], there is a finite subcollection V1 , . . . , Vn of the V ’s which cover [0, 1]. Let W1 , . . . , Wn be the corresponding W ’s in the family Φb and let Ub = W1 ∩ · · · ∩ Wn . But then E|Ub × Wi , i = 1, . . . , n are all trivial and thus by Lemma 3.4.32, E|Ub × [0, 1] is trivial. Then { Ub | b ∈ B } is the desired open covering of B.  3.4.34 Lemma. trivial.

Let π : E → B × [0, 1] be a C 0 fiber bundle such that E|B × {0} is trivial. Then E is

Proof. By Lemma 3.4.33, there is an open cover {Ui } of B such that E|Ui × [0, 1] is trivial; let ϕi be the corresponding trivializations. Denote by ϕ : E|B × {0} → B × {0} × F the trivialization guaranteed in the hypothesis of the lemma, where F is the typical fiber of E. We modify all ϕi in such a way that ϕi : E|Ui × {0} → Ui × {0} × F coincides with ϕ : E|Ui × {0} → Ui × {0} × F in the following way. The homeomorphism ϕi ◦ ϕ−1 : Ui × {0} × F → Ui × {0} × F is of the form (b, 0, f ) 7→ (b, 0, αbi (f )) for αbi : F → F a homeomorphism depending continuously on b ∈ B. Define χi : Ui × [0, 1] × F → Ui × [0, 1] × F by χi (b, s, f ) = (b, s, (αbi )−1 (f )). Then ψi = χi ◦ ϕi : E|Ui × [0, 1] → Ui × [0, 1] × F maps any e ∈ π −1 (B × {0}) to ϕ(e). Assume each ϕi on E|Ui × {0} equals ϕ on E|Ui × {0}. Define λi : E|Ui × [0, 1] → Ui × {0} × F to be the composition of the map (b, s, f ) ∈ Ui × [0, 1] × F 7→ (b, 0, f ) ∈ Ui × {0} × F with ϕi . Since each ϕi coincides with ϕ on E|Ui × {0}, it follows that whenever Ui ∩ Uj 6= ∅, λi and λj coincide on E|(Ui ∩ Uj ) × [0, 1], so that the collection of all {λi } define a fiber bundle map λ : E → B × {0} × F over the map χ : (b, s) ∈ B × [0, 1] 7→ (b, 0) ∈ B × {0}. By the fiber bundle version 3.4-23 of Exercise 3.4-15(i) and (iii), E equals the pull-back χ∗ (B × {0} × F ). Since the bundle B × {0} × F → B × {0} is trivial, so is its pull-back E.  3.4.35 Theorem.

Let π : E → B be any C 0 fiber bundle over a contractible space B. Then E is trivial.

Proof. By hypothesis, there is a homotopy h : B × [0, 1] → B such that h(b, 0) = b0 and h(b, 1) = b for any b ∈ B, where b0 ∈ B is a fixed element of B. Then the pull-back bundle h∗ E is a fiber bundle over B × [0, 1] whose restrictions to B × {0} and B × {1} equals the trivial fiber bundle over {b0 } and E over B × {1}, respectively. By Lemma 3.4.34, h∗ E is trivial over B × [0, 1] and thus E, which is isomorphic to E|B × {1}, is also trivial.  All previous proofs go through without any modifications to the C k -case, once manifolds with boundary are defined (see §7.1).

Exercises  3.4-1. Let N ⊂ M be a submanifold. Show that T N is a subbundle of T M |N and thus is a submanifold of T M .  3.4-2. Find an explicit example of a fiber-preserving diffeomorphism between vector bundles that is not a vector bundle isomorphism.

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 3.4-3.

Let ρ : R × S n → S n and σ : Rn+1 × S n → S n be trivial vector bundles. Show that T S n ⊕ (R × S n ) ∼ = (Rn+1 × S n ).

Hint : Realize ρ as the vector bundle whose one-dimensional fiber is the normal to the sphere.  3.4-4. (i) Let π : E → B be a vector bundle. Show that T E|B is vector bundle isomorphic to E ⊕ T B. Conclude that E is isomorphic to a subbundle of T E. Tπ

Hint: The short exact sequence 0 → E → T E|B → T B → 0 splits via T i, where i : B → E is the inclusion of B as the zero section of E; apply Proposition 3.4.26. (ii) Show that the isomorphism ϕE found in (i) is natural, that is, if π 0 : E 0 → B 0 is another vector bundle, f : E → E 0 is a vector bundle map over fB : B → B 0 , and ϕE 0 : T E 0 |B → E 0 ⊕ T B 0 is the isomorphism in (i) for π 0 : E 0 → B 0 , then ϕE 0 ◦ T f = (f ⊕ T fB ) ◦ ϕE .  3.4-5. Show that the mapping s : E ⊕ E → E, s(e, e0 ) = e + e0 (fiberwise addition) is a vector bundle mapping over the identity.  3.4-6.

Write down explicitly the charts in Examples 3.4.25 given by Proposition 3.4.24.

 3.4-7. (i) A vector bundle π : E → B is called stable if its Whitney sum with a trivial bundle over B is trivial. Show that T S n is stable, but the M¨obius band M is not. (ii) Two vector bundles π : E → B, ρ : E → B are called stably isomorphic if the Whitney sum of E with some trivial bundle over B is isomorphic with the Whitney sum of F with (possibly another) trivial vector bundle over B. Let KB be the set of stable isomorphism classes of vector bundles with finite dimensional fiber over B. Show that the operations of Whitney sum and of tensor product induce on KB a ring structure. Find a surjective ring homomorphism of KB onto Z.  3.4-8. A vector bundle with one-dimensional fibers is called a line bundle. Show that any line bundle which admits a global nowhere vanishing section is trivial.  3.4-9. Generalize Example 3.4.25B S to vector bundles with different bases. If π : E → M and ρ : F → N are vector bundles, show that the set (m,n)∈M ×N L(Em , Fn ) is a vector bundle with base M × N . Describe the fiber and compute the relevant dimensions in the finite dimensional case.  3.4-10. Let N be a submanifold of M . The normal bundle ν(N ) of N is defined to be ν(N ) = (T M |N )/T N . Assume that N has finite codimension k. Show that ν(N ) is trivial iff there are smooth maps Xi : N → T M , i = 1, . . . , k such that Xi (n) ∈ Tn M and { Xi (n) | i = 1, . . . , k } span a subspace Vn satisfying Tn M = Tn N ⊕ Vn for all n ∈ N . Show that ν(S n ) is trivial.  3.4-11. Let N be a submanifold of M . Prove that the conormal bundle defined by µ(N ) = { α ∈ Tn∗ M | hα, ui = 0 for all u ∈ Tn N and all n ∈ N } in a subbundle of T ∗ M |N which is isomorphic to the normal bundle ν(N ) defined in Exercise 3.4-10. Generalize the constructions and statements of 3.4-10 and the current exercise to an arbitrary vector subbundle F of a vector bundle E.  3.4-12. (i) Use the fact that S 3 is the unit sphere in the associative division algebra H to show that 3 T S is trivial. (ii) Cayley numbers. Consider on R8 = H ⊕ H the usual Euclidean inner product h , i and define a multiplication in R8 by (a1 , b1 )(a2 , b2 ) = (a1 a2 − b∗2 b1 , b2 a1 + b1 a∗2 ) where ai , bi ∈ H, i = 1, 2, and the multiplication on the right hand side is in H. Prove the relation hα1 β1 , α2 β2 i + hα2 β1 , α1 β2 i = 2 hα1 , α2 i hβ1 , β2 i where hα, βi denotes the dot product in R8 . Show that if one defines the conjugate of (a, b) by (a, b)∗ = (a∗ , −b), then k(a, b)k2 = (a, b)[(a, b)∗ ]. Prove that kαβk = kαk kβk for all α, β ∈ R8 . Use this relation to show that R8 is a nonassociative division algebra over R, C, and H. R8 with this algebraic structure is called the algebra of Cayley numbers or algebra of octaves; it is denoted by O.

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173

(iii) Show that O is generated by 1 and seven symbols e1 , . . . , e7 satisfying the relations e2i = −1, ei ej = −ej ei , e1 e6 = −e7 , e2 e5 = e7 ,

e1 e2 = e3 , e2 e4 = e6 ,

e1 e4 = e5 , e3 e4 = e7 ,

e3 e5 = −e6 ,

together with 14 additional relations obtained by cyclic permutations of the indices in the last 7 relations. Hint: The isomorphism is given by associating 1 to the element (1, 0, . . . , 0) ∈ R8 and to ei the vector in R8 having all entries zero with the exception of the (i + 1)st which is 1. (iv) Show that any two elements of O generate an associative algebra isomorphic to a subalgebra of H. Hint: Show that any element of O is of the form a + be4 for a, b ∈ H. (v) Since S 7 is the unit sphere in O, show that T S 7 is trivial.  3.4-13. (i) Let π : E → B be a locally trivial fiber bundle. Show that V = ker(T π) is a vector subbundle of T E, called the vertical bundle. A vector subbundle H of T E such that V ⊕ H = T E is called a horizontal subbundle. Show that T π induces a vector bundle map H → T E over π which is an isomorphism on each fiber. (ii) If π : E → M is a vector bundle, show that each fiber Vv of V , v ∈ E is naturally identified with Eb , where b = π(v). Show that there is a natural isomorphism of T0 E with Tb B ⊕ Eb , where 0 is the zero vector in Eb . Argue that there is in general no such natural isomorphism of Tv E for v 6= 0.  3.4-14.

Let En be the trivial vector bundle RPn × Rn+1 .

(i) Show that Fn = { ([x], λx) | x ∈ Rn+1 , λ ∈ R } is a line subbundle of En . Hint: Define n

f : En → RP × R

n+1

 by f ([x], y) =

x [x], y − (x · y) kxk2



and show that f is a vector bundle map having the restriction to each fiber a linear map of rank n. Apply Proposition 3.4.18. (ii) Show that Fn is isomorphic to γ1 (Rn+1 ). (iii) Show that F1 is isomorphic to the M¨obius band M. (iv) Show that Fn is the quotient bundle of the normal bundle ν(S n ) to S n by the equivalence relation which identifies antipodal points and takes the outward normal to the inward normal. Show that the projection map ν(S n ) → Fn is a 2 to 1 covering map. (v) Show that Fn is nontrivial for all n ≥ 1. Hint: Use (iv) to show that any section σ of Fn vanishes somewhere; do this by considering the associated section σ ∗ of the trivial normal bundle to S n and using the intermediate value theorem. (vi) Show that any line bundle over S 1 is either isomorphic to the cylinder S 1 × R or the M¨obius band M.  3.4-15. (i) Let π : E → B be a vector bundle and f : B 0 → B a smooth map. Define the pull-back bundle f ∗ π : f ∗ E → B 0 by f ∗ E = { (v, b0 ) | π(v) = f (b0 ) },

f ∗ π(v, b0 ) = b0

and show that it is a vector bundle over B 0 , whose fibers over b0 equal Ef (b0 ) . Show that h : f ∗ E → E, h(e, b) = e, is a vector bundle map which is the identity on every fiber. Show that the pull-back bundle of a trivial bundle is trivial.

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(ii) If g : B 00 → B 0 show that (f ◦g)∗ π : (f ◦g)∗ E → B 00 is isomorphic to the bundles g ∗ f ∗ π : g ∗ f ∗ E → B 00 . Show that isomorphic vector bundles have isomorphic pull-backs. (iii) If ρ : E 0 → B 0 is a vector bundle and g : E 0 → E is a vector bundle map inducing the map f : B 0 → B on the zero sections, then prove there exists a unique vector bundle map g ∗ : E 0 → f ∗ E inducing the identity on B 0 and is such that h ◦ g ∗ = g. (iv) Let σ : F → B be a vector bundle and u : F → E be a vector bundle map inducing the identity on B. Show that there exists a unique vector bundle map f ∗ u : f ∗ F → f ∗ E inducing the identity on B 0 and making the diagram f ∗u

F

? F

- f ∗E

? - E u

(v) If π : E → B, π 0 : E 0 → B are vector bundles and if ∆ : B → B × B is the diagonal map b 7→ (b, b), show that E ⊕ E 0 ∼ = ∆∗ (E × E 0 ). (vi) Let π : E → B and π 0 : E 0 → B be vector bundles and denote by π : B × B → B, i = 1, 2 the projections. Show that E × E 0 ∼ = p∗1 (E) ⊕ p∗2 (E 0 ) and that the following sequences are split exact: 0 → E → E ⊕ E 0 → E 0 → 0. 0 → E 0 → E ⊕ E 0 → E → 0. 0 → p∗1 (E) → E × E 0 → p∗2 (E 0 ) → 0. 0 → p∗2 (E 0 ) → E × E 0 → p∗1 (E) → 0.  3.4-16. (i) Show that Gk (Rn ) is a submanifold of Gk+1 (Rn+1 ). Denote by i : Gk (Rn ) → Gk+1 (Rn+1 ), i(F ) = F × R the canonical inclusion map. (ii) If ρ : R × Gk (Rn ) → Gk (Rn ) is the trivial bundle, show that the pull-back bundle i∗ (γk+1 (Rn+1 )) is isomorphic to γk (Rn ) ⊕ (R × Gk (Rn )).  3.4-17.

Show that T (M1 × M2 ) ∼ = p∗1 (T M1 ) ⊕ p∗2 (T M2 )

where pi : M1 × M2 → Mi , i = 1, 2 are the canonical projections and p∗i (T Mi ) denotes the pull-back bundle defined in Exercise 3.4-15.  3.4-18.

(i) Let π : E → B be a vector bundle. Show that there is a short exact sequence f

g

0 −→ π ∗ E −→ T −→ E −→ π ∗ (T B) −→ 0 where

d f (v, v ) = (v + tv 0 ) dt t=0 0

and

g(vv ) = (Tv π(uv ), π(v))

(ii) Show that ker(T π) is a subbundle of T E, called the vertical subbundle of T E. Any subbundle H ⊂ T E such that T E = ker(T π) ⊕ H, is called a horizontal subbundle of T E. Show that T π induces an isomorphism of H with π ∗ (T B).

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175

(iii) Show that if T E admits a horizontal subbundle then the sequence in (i) splits.  3.4-19. Let π : E → B, ρ : F → C be vector bundles and let f : B → C be a smooth map. Define Lf (E, F ) = Γ∗f L(E, F ), where Γf : b ∈ B 7→ (b, f (b)) ∈ B × C is the graph map defined by f . Show that sections of Lf (E, F ) coincide vector bundle maps E → F over f .  3.4-20.

Let M be an n-manifold. A frame at m ∈ M is an isomorphism α : Tm M → Rn . Let F(M ) = { (m, α) | α is a frame at m }.

Define π : F(M ) → M by π(m, α) = m. (i) Let (U, ϕ) be a chart on M . Show that (m, α) ∈ π −1 (U ) 7→ (m, Tm ϕ ◦ α−1 ) ∈ U × GL(Rn ) is a diffeomorphism. Prove that these diffeomorphisms as (U, ϕ) vary over a maximal atlas of M define by collation a manifold structure on F(M ). Prove that π : F(M ) → M is a locally trivial fiber bundle with typical fiber GL(n). (ii) Prove that the sequence i

π∗ τ

0 −→ ker(T π) −→ T F(M ) −→ π ∗ (T M ) −→ 0 is short exact, where i is the inclusion and π ∗ τ is the vector bundle projection π ∗ (T M ) → F(M ) induced by the tangent bundle projection τ : T M → M . (iii) Show that ker(T π) and π ∗ (T M ) are trivial vector bundles. h

(iv) A splitting 0 → π ∗ (T M ) → T F(M ) of the short sequence in (ii) is called a connection on M . Show that if M has a connection, then T F(M ) = ker(T π) ⊕ H, where H is a subbundle of T F(M ) whose fibers are isomorphic by T π to the fiber of T M .  3.4-21. (i) Generalize the Hopf fibration to the complex Hopf fibrations κn : S 2n+1 → CPn with fiber S 1 . (ii) Replace in (i) C by the division algebra of quaternions H. Generalize (i) to the quaternionic Hopf fibrations χn : S 4n+3 → HPn with fiber S 3 . HPn is the quaternionic space defined as the set of one dimensional vector subspaces over H in Hn+1 . Is anything special happening when n = 1? Describe.  3.4-22. (i) Try to define OPn , where O are the Cayley numbers. Show that the proof of transitivity of the equivalence relation in On+1 requires associativity. (ii) Define p0 (a, b) = ab−1 if b 6= 0 and p0 (a, b) = ∞ if b = 0, where S 8 is thought of as the one-point compactification of R8 = O (see Exercise 3.1-5). Show p0 is smooth and prove that p = p0 |S 15 is onto. Proceed as in Example 3.4.29D and show that p : S 15 → S 8 is a fiber bundle with typical fiber S 7 whose bundle structure is given by an atlas with two fiber bundle charts.  3.4-23.

3.5

Define the pull-back of fiber bundles and prove properties analogous to those in Exercise 3.4-15.

Submersions, Immersions, and Transversality

The notions of submersion, immersion, and transversality are geometric ways of stating various hypotheses needed for the inverse function theorem, and are central to large portions of calculus on manifolds. One immediate benefit is easy proofs that various subsets of manifolds are actually submanifolds. 3.5.1 Theorem (Local Diffeomorphisms Theorem). Suppose that M and N are manifolds, f : M → N is of class C r , r ≥ 1 and m ∈ M . Suppose T f restricted to the fiber over m ∈ M is an isomorphism. Then f is a C r diffeomorphism from some neighborhood of m onto some neighborhood of f (m).

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Proof. In local charts, the hypothesis reads: (Dfϕψ )(u) is an isomorphism, where ϕ(m) = u. Then the inverse function theorem guarantees that fϕψ restricted to a neighborhood of u is a C r diffeomorphism. Composing with chart maps gives the result.  The local results of Theorems 2.5.9 and 2.5.13 give the following: 3.5.2 Theorem (Local Onto Theorem). Let M and N be manifolds and f : M → N be of class C r , where r ≥ 1. Suppose T f restricted to the fiber Tm M is surjective to Tf (m) N . Then (i) f is locally onto at m; that is, there are neighborhoods U of m and V of f (m) such that f |U : U → V is onto; in particular, if T f is surjective on each tangent space, then f is an open mapping; (ii) if, in addition, the kernel ker(Tm f ) is split in Tm M there are charts (U, ϕ) and (V, ψ) with m ∈ U , f (U ) ⊂ V , ϕ : U → U 0 × V 0 , ϕ(m) = (0, 0), ψ : V → V 0 , and fϕψ : U 0 × V → V 0 is the projection onto the second factor. Proof.

It suffices to prove the results locally, and these follow from Theorems 2.5.9 and 2.5.13.



Submersions. The notions of submersion and immersion correspond to the local surjectivity and injectivity theorems from §2.5. Let us first examine submersions, building on the preceding theorem. 3.5.3 Definition. Suppose M and N are manifolds with f : M → N of class C r , r ≥ 1. A point n ∈ N is called a regular value of f if for each m ∈ f −1 ({n}), Tm f is surjective with split kernel. Let Rf denote the set of regular values of f : M → N ; note N \f (M ) ⊂ Rf ⊂ N . If, for each m in a set S, Tm f is surjective with split kernel, we say f is a submersion on S. Thus n ∈ Rf iff f is a submersion on f −1 ({n}). If Tm f is not surjective, m ∈ M is called a critical point and n = f (m) ∈ N a critical value of f . 3.5.4 Theorem (Submersion Theorem).

Let f : M → N be of class C ∞ and n ∈ Rf . Then the level set

f −1 (n) = { m | m ∈ M, f (m) = n } is a closed submanifold of M with tangent space given by Tm f −1 (n) = ker Tm f . Proof. If B is a submanifold of M and b ∈ B, we need to clarify in what sense Tb B is a subspace of Tb M . Letting i : B → M be the inclusion, Tb i : Tb B → Tb M is injective with closed split range. Hence Tb B can be identified with a closed split subspace of Tb M . If f −1 (n) = ∅ the theorem is clearly valid. Otherwise, for m ∈ f −1 (n) we find charts (U, ϕ), (V, ψ) as described in Theorem 3.5.2. Because −1 ϕ(U ∩ f −1 (n)) = fϕψ (0) = U 0 × {0},

we get the submanifold property. (See Figure 3.5.1.) Since fϕψ : U 0 × V 0 → V 0 is the projection onto the second factor, where U 0 ⊂ E and V 0 ⊂ F, we have −1 Tu (fϕψ (0)) = Tu U 0 = E = ker(Tu fϕψ )

which is the local version of the second statement.

for u ∈ U 0 , 

If N is finite dimensional and n ∈ Rf , observe that codim(f −1 (n)) = dim N , from the second statement of Theorem 3.5.4. (This makes sense even if M is infinite dimensional.) Sard’s theorem, discussed in the next section, implies that Rf is dense in N . 3.5.5 Examples.

3.5 Submersions, Immersions, and Transversality

f

177

n

Figure 3.5.1. Submersion theorem

A. We shall use the preceding theorem to show that S n ⊂ Rn+1 is a submanifold. Indeed, let f : Rn+1 → R be defined by f (x) = kxk2 , so S n = f −1 (1). To show that S n is a submanifold, it suffices to show that 1 is a regular value of f . Suppose f (x) = 1. Identifying T Rn+1 = Rn+1 × Rn+1 , and the fiber over x with elements of the second factor, we get (Tx f )(v) = Df (x) · v = 2 hx, vi . Since x 6= 0, this linear map is not zero, so as the range is one-dimensional, it is surjective. The same argument shows that the unit sphere in Hilbert space is a submanifold. B. Stiefel Manifolds.

Define

St(m, n; k) = { A ∈ L(Rm , Rn ) | rank A = k },

where k ≤ min(m, n).

Using the preceding theorem we shall prove that St(m, n; k) is a submanifold of L(Rm , Rn ) of codimension (m−k)(n−k); this manifold is called the Stiefel manifold and plays an important role in the study of principal fiber bundles. To show that St(m, n; k) is a submanifold, we will prove that every point A ∈ St(m, n; k) has an open neighborhood U in L(Rm , Rn ) such that St(m, n; k) ∩ U is a submanifold in L(Rm , Rn ) of the right codimension; since the differentiable structures on intersections given by two such U coincide (being induced from the manifold structure of L(Rm , Rn )), the submanifold structure of St(m, n; k) is obtained by collation (Exercise 3.2-6). Let A ∈ St(m, n; k) and choose bases of Rm , Rn such that   a b A= c d with a an invertible k × k matrix. The set    x y U= x is an invertible k × k matrix z v is open in L(Rn , Rn ). An element of U has rank k iff v − zx−1 y = 0. Indeed   I 0 [8pt] − zx−1 I

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is invertible and 

I [8pt] − zx−1

so rank

# " # " x y 0 x y = , I z v 0 v − zx−1 y

   x y x = rank z v 0

y v − zx−1 y



equals k iff v − zx−1 y = 0. Define f : U → L(Rm−k , Rn−k ) by   x y f = v − zx−1 y. z v The preceding remark shows that f −1 (0) = St(m, n; k) ∩ U and thus if f is a submersion, f −1 (0) is a submanifold of L(Rm , Rn ) of codimension equal to dim L(Rm−k , Rn−k ) = (m − k)(n − k). To see that f is a submersion, note that for x, y, z fixed, the map v 7→ v − zx−1 y is a diffeomorphism of L(Rm−k , Rn−k ) to itself. C. Orthogonal Group. Let O(n) be the set of elements Q of L(Rn , Rn ) that are orthogonal, that is, QQT = Identity. We shall prove that O(n) is a compact submanifold of dimension n(n − 1)/2. This manifold is called the orthogonal group of Rn ; the group operations (composition of linear operators and inversion) being smooth in L(Rn , Rn ) are therefore smooth in O(n), that is, O(n) is an example of a Lie group. To show that O(n) is a submanifold, let sym(n) denote the vector space of symmetric linear operators S of Rn , that is, ST = S; its dimension equals n(n + 1)/2. The map f : L(Rn , Rn ) → sym(n), f (Q) = QQT is smooth and has derivative TQ f (A) = AQT + QAT = AQ−1 + QAT at Q ∈ O(n). This linear map from L(Rn , Rn ) to sym(n) is onto since for any S ∈ sym(n), TQ f (SQ/2) = S. Therefore, by Theorem 3.5.4, f −1 (Identity) = O(n) is a closed submanifold of L(Rn , Rn ) of dimension equal to n2 − n(n + 1)/2 = n(n − 1)/2. Finally, O(n) is compact since it lies on the unit sphere of L(Rn , Rn ). D. Orthogonal Stiefel Manifold.

Let k ≤ n and

Fk,n = OSt(n, n; k) = { orthonormal k-tuples of vectors in Rn }. We shall prove that OSt(n, n; k) is a compact submanifold of O(n) of dimension nk − k(k + 1)/2; it is called the orthogonal Stiefel manifold . Any n-tuple of orthonormal vectors in Rn is obtained from the standard basis e1 , . . . , en of Rn by an orthogonal transformation. Since any k-tuple of orthonormal vectors can be completed via the Gram–Schmidt procedure to an orthonormal basis, the set OSt(n, n; k) equals f −1 (0), where f : O(n) → O(n − k) is given by letting f (Q) = Q0 , where Q0 = the (n − k) × (n − k) matrix obtained from Q by removing its first k rows and columns. Since TQ f (A) = A0 is onto, it follows that f is a submersion. Therefore, f −1 (0) is a closed submanifold of O(n) of dimension equal to k(k + 1) n(n − 1) (n − k)(n − k − 1) − = nk − . 2 2 2



3.5 Submersions, Immersions, and Transversality

Immersions.

179

Now we look at maps whose derivatives are one-to-one.

3.5.6 Definition. A C r map f : M → N , r ≥ 1, is called an immersion at m if Tm f is injective with closed split range in Tf (m) N . If f is an immersion at each m, we just say f is an immersion. 3.5.7 Theorem (Immersion Theorem). equivalent:

For a C r map f : M → N , where r ≥ 1, the following are

(i) f is an immersion at m; (ii) there are charts (U, ϕ) and (V, ψ) with m ∈ U , f (U ) ⊂ V , ϕ : U → U 0 , ψ : V → U 0 × V 0 and ϕ(m) = 0 such that fϕψ : U 0 → U 0 × V 0 is the inclusion u 7→ (u, 0); (iii) there is a neighborhood U of m such that f (U ) is a submanifold in N and f restricted to U is a diffeomorphism of U onto f (U ). Proof. The equivalence of (i) and (ii) is guaranteed by the local immersion theorem 2.5.12. Assuming (ii), choose U and V given by that theorem to conclude that f (U ) is a submanifold in V . But V is open in N and hence f (U ) is a submanifold in N proving (iii). The converse is a direct application of the definition of a submanifold.  It should be noted that the theorem does not imply that f (M ) is a submanifold in N . For example f : S 1 → R2 , given in polar coordinates by r = cos 2θ, is easily seen to be an immersion (by computing T f using the curve c(θ) = cos(2θ) on S 1 but f (S 1 ) is not a submanifold of R2 : any neighborhood of 0 in R2 intersects f (S 1 ) in a set with corners” which is not diffeomorphic to an open interval. In such cases we say f is an immersion with self-intersections. See Figure 3.5.2.

f

y

x

Figure 3.5.2. Images of immersions need not be submanifolds

In the preceding example f is not injective. But even if f is an injective immersion, f (M ) need not be a submanifold of N , as the following example shows. Let f be a curve whose image is as shown in Figure 3.5.3. Again the problem is at the origin: any neighborhood of zero does not have the relative topology given by N . Embeddings. N.

If f : M → N is an injective immersion, f (M ) is called an immersed submanifold of

3.5.8 Definition. An immersion f : M → N that is a homeomorphism onto f (M ) with the relative topology induced from N is called an embedding . Thus, if f : M → N is an embedding, then f (M ) is a submanifold of N . The following is an important situation in which an immersion is guaranteed to be an embedding; the proof is a straightforward application of the definition of relative topology.

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y f

x

Figure 3.5.3. Images of injective immersions need not be submanifolds

3.5.9 Theorem (Embedding Theorem). image is an embedding.

An injective immersion which is an open or closed map onto its

With regard to the hypothesis in this theorem, we note that the condition “f : M → N is closed” is implied by “f is proper.” To show this, first recall from Exercise 1.5-10 that “proper” in the context of manifolds means that each sequence xn ∈ M with the property that f (xn ) is convergent to a point y ∈ N has a convergent subsequence in M . Now we must show that if A is a closed subset of M , then f (A) is closed in N . To do this, let xn ∈ A, and suppose f (xn ) = yn converges to y ∈ N . Then there is a subsequence {zm } of {xn }, such that zm → x. Since A is closed, x ∈ A and by continuity of f , y = f (x) ∈ f (A); that is, f (A) is closed. The reader should note that while both hypotheses in the proposition are necessary, properness of f is only sufficient. An injective nonproper immersion whose image is a submanifold is, for example, the map f : ]0, ∞[ → R2 given by   1 1 f (t) = t cos , t sin . t t This map f is an open map onto its image and so Theorem 3.5.9 applies; the submanifold f (]0, ∞[) is a spiral around the origin. Transversality.

This is an important notion that applies to both maps and submanifolds.

3.5.10 Definition. A C r map f : M → N , r ≥ 1, is said to be transversal to the submanifold P of N (denoted f t P ) if either f −1 (P ) = ∅, or if for every m ∈ f −1 (P ), T1. (Tm f )(Tm M ) + Tf (m) P = Tf (m) N and T2. the inverse image (Tm f )−1 (Tf (m) P ) of Tf (m) P splits in Tm M . The first condition T1 is purely algebraic; no splitting assumptions are made on (Tm f )(Tm M ), nor need the sum be direct. If M is a Hilbert manifold, or if M is finite dimensional, then the splitting condition T2 in the definition is automatically satisfied. 3.5.11 Examples. A.

If each point of P is a regular value of f , then f t P since, in this case, (Tm f )(Tm M ) = Tf (m) N .

B. Assume that M and N are finite-dimensional manifolds with dim(P ) + dim(M ) < dim(N ). Then f t P implies f (M ) ∩ P = ∅. This is seen by a dimension count: if there were a point m ∈ f −1 (P ) ∩ M , then dim(N ) = dim((Tm f )(Tm M ) + Tf (m) P ) ≤ dim(M ) + dim(P ) < dim(N ), which is absurd.

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181

C. Let M = R2 , N = R3 , P = the (x, y) plane in R3 , a ∈ R and define fa : M → N , by fa (x, y) = (x, y, x2 + y 2 + a). Then f t P if a 6= 0; see Figure 3.5.4. This example also shows intuitively that if a map is not transversal to a submanifold it can be perturbed very slightly to a transversal map; for a discussion of this phenomenon we refer to the Supplement 3.6B. 

y

y x

x

x

Figure 3.5.4. These manifolds are nontransverse to the xy plane at a = 0.

3.5.12 Theorem (Transversal Mapping Theorem). Let f : M → N be a C ∞ map and P a submanifold of N . If f t P , then f −1 (P ) is a submanifold of M and Tm (f −1 (P )) = (Tm f )−1 (Tf (m) P ) for all m ∈ f −1 (P ). If P has finite codimension in N , then codim(f −1 (P )) = codim(P ). Proof.

Let (V, ψ) be a chart at f (m0 ) ∈ P in N with the submanifold property for P ; let ψ(V ) = V1 × V2 ⊂ F1 ⊕ F2 ,

ψ(V ∩ P ) = V1 × {0},

ψ(f (m0 )) = (0, 0)

and denote by p2 : V1 × V2 → V2 the canonical projection. Let (U, ϕ) be a chart at m0 in M , such that ϕ(m0 ) = 0, ϕ : U → ϕ(U ) ⊂ E and f (U ) ⊂ V . For m ∈ U ∩ f −1 (P ), Tm (p2 ◦ ψ ◦ f |U ) = p2 ◦ Tf (m) ψ ◦ Tm f and Tm (ψ ◦ f )(Tm M ) + F1 = F1 ⊕ F2 (by transversality of f on P ). Hence Tm (p2 ◦ ψ ◦ f |U ) : Tm U = Tm M → F2 is onto. Its kernel is (Tm f )−1 (Tf (m) P ) since ker p2 = F1 and (Tf (m) ψ)−1 (F1 ) = Tf (m) P, and thus it is split in Tm M . In other words, 0 is a regular value of p2 ◦ ψ ◦ f |U : U → F2 and thus (p2 ◦ ψ ◦ f |U )−1 (0) = f −1 (P ∩ V ) is a submanifold of U , and hence of M whose tangent space at m ∈ U equals ker(Tm (p2 ◦ ψ ◦ f |U )) = (Tm f )−1 (Tf (m) P ) by the submersion theorem 3.5.4. Thus f −1 (P ∩ V ) is a submanifold of M for any chart domain V with the submanifold property; that is, f −1 (P ) is a submanifold of M . If P has finite codimension then F2 is finite dimensional and thus again by the submersion theorem, codim(f −1 (P )) = codim f −1 (P ∩ V ) = dim(F2 ) = codim(P ).



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3. Manifolds and Vector Bundles

Notice that this theorem reduces to the submersion theorem if P is a point. 3.5.13 Corollary. Suppose that M1 and M2 are submanifolds of M , m ∈ M1 ∩M2 , Tm M1 +Tm M2 = Tm M , and that Tm M1 ∩ Tm M2 splits in Tm M for all m ∈ M1 ∩ M2 ; this condition is denoted M1 t M2 and we say M1 and M2 are transversal . Then M1 ∩ M2 is a submanifold of M and Tm (M1 ∩ M2 ) = Tm M1 ∩ Tm M2 . M1 and M2 are said to intersect cleanly when this conclusion holds. (Transversality thus implies clean intersection.) If both M1 and M2 have finite codimension in M , then codim(M1 ∩ M2 ) = codim(M1 ) + codim(M2 ). Proof. The inclusion i1 : M1 → M , satisfies i1 t M2 , and i−1 1 (M2 ) = M1 ∩ M2 . Now apply the previous theorem.  3.5.14 Examples. A. In R3 , the unit sphere M1 = { (x, y, z) | x2 + y 2 + z 2 = 1 } intersects the cylinder M2 = { (x, y, z) | x2 + y 2 = a } transversally if 0 < a 6= 1; M1 ∩ M2 = ∅ if a > 1 and M1 ∩ M2 is the union of two circles if 0 < a < 1 (Figure 3.5.5).

Figure 3.5.5. A sphere intersects a cylinder transversally except at the critical value a = 1

B. The twisted ribbon M1 in Figure 3.5.6 does not meet M2 , the xy plane, in a manifold, so M1 is not transversal to M2 . C. Let M be the xy-plane in R3 and N be the graph of f (x, y) = (xy)2 . Even though T N ∩ T M has constant dimension (equal to 2), N ∩ M is not a manifold (Figure 3.5.7).  Subimmersions. There is one more notion connected with geometric ways to state the implicit function theorem that generalizes submersions in a different way than transversality. Roughly speaking, instead of requiring that a map f : M → N have onto tangent map at each point, one asks that its rank be constant. 3.5.15 Definition. A C r map f : M → N , r ≥ 1 is called a subimmersion if for each point m ∈ M there is an open neighborhood U of m, a manifold P , a submersion s : U → P , and an immersion j : P → N such that f |U = j ◦ s. Note that submersions and immersions are subimmersions. The order submersion followed by immersion in this definition is important, because the opposite order would yield nothing. Indeed, if f : M → N is any C ∞ mapping, then we can write f = p2 ◦ j, where j : M → M × N , is given by j(m) = (m, f (m)) and p2 : M × N → N the canonical projection. Clearly j is an immersion and p2 a submersion, so any mapping can be written as an immersion followed by a submersion.

3.5 Submersions, Immersions, and Transversality

183

Figure 3.5.6. The twisted ribbon

Figure 3.5.7. Constancy of dimension of T M ∩ T N does not imply smoothness of M ∩ N

The following will connect the notion of subimmersion to that of “constant rank.” 3.5.16 Proposition. (i) A C ∞ map f : M → N is a subimmersion iff for every m ∈ M there is a chart (U, ϕ) at m, where ϕ : U → V1 × U2 ⊂ F1 ⊕ E2 ,

ϕ(m) = (0, 0),

and a chart (V, ψ) at f (m) where f (U ) ⊂ V,

ψ : V → V1 × V2 ⊂ F1 ⊕ F2 ,

ψ(f (m)) = (0, 0)

such that fϕψ (x, y) = (x, 0). (ii) If M or N are finite dimensional, f is a subimmersion iff the rank of the linear map Tm f : Tm M → Tf (m) N is constant for m in each connected component of M . Proof. (i) follows from Theorem 3.5.2, Theorem 3.5.7(ii) and the composite mapping theorem; alternatively, one can use Theorem 2.5.15. If M or N are finite dimensional, then necessarily rank(Tm f ) is finite and thus by (i) the local representative Tm f has constant rank in a chart at m; that is, the rank of Tm f is constant on connected components of M , thus proving (ii). The converse follows by Theorem 2.5.15 and (i). 

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3.5.17 Theorem (Subimmersion Theorem). Suppose f : M → N is C ∞ , n0 ∈ N and f is a subimmersion in an open neighborhood of f −1 (n0 ). (If M or N are finite dimensional this is equivalent to Tm f having constant rank in a neighborhood of each m ∈ f −1 (n0 ).) Then f −1 (n0 ) is a submanifold of M with Tm f −1 (n0 ) = ker(Tm f ). Proof. If f −1 (n0 ) = ∅ there is nothing to prove. If m ∈ f −1 (n0 ) find charts (U, ϕ) at m and (V, ψ) at −1 f (m) = n0 given by Proposition 3.5.16(i). Since ϕ(U ∩ f −1 (n0 )) = fϕψ (0) = {0} × U2 we see that (U, ϕ) has −1 the submanifold property for f (n0 ). In addition, if u ∈ U2 , then −1 T(0,u) (fϕψ (0)) = Tu U2 = E2 = ker(T(0,u) fϕψ ),

which is the local version of the second statement.



Notice that codim(f −1 (n0 )) = rank(Tm f )

for m ∈ f −1 (n0 )

if rank(Tm f ) is finite. The subimmersion theorem reduces to Theorem 3.5.4 when f is a submersion. The immersion part of the subimmersion f implies a version of Theorem 3.5.7(iii). 3.5.18 Theorem (Fibration Theorem).

The following are equivalent for a C ∞ map f : M → N :

(i) the map f is a subimmersion (if M or N are finite dimensional, this is equivalent to the rank of Tm f being locally constant); (ii) for each m ∈ M there is a neighborhood U of m, a neighborhood V of f (m), and a submanifold Z of M with m ∈ Z such that f (U ) is a submanifold of N and f induces a diffeomorphism of f −1 (V ) ∩ Z ∩ U onto f (U ) ∩ V ; (iii) ker T f is a subbundle of T M (called the tangent bundle to the fibers) and for each m ∈ M , the image of Tm f is closed and splits in Tf (m) N . If one of these hold and if f is open (or closed ) onto its image, then f (M ) is a submanifold of N . Proof. To show that (i) implies (ii), choose for U and V the chart domains given by Proposition 3.5.16(i) and let Z = ϕ−1 (V1 × {0}). Then ψ(f (U ) ∩ V ) = fϕψ (V1 × U2 ) = V1 × {0}, that is, f (U ) is a submanifold of N . In addition, the local expression of f : f −1 (V ) ∩ Z ∩ U → f (U ) ∩ V is (x, 0) 7→ (x, 0), thus proving that so restricted, f is a diffeomorphism. Reading this argument backward shows that (ii) implies (i). We have T ϕ(ker Tf ∩ TU ) = V1 × U2 × {0} × E2 , since fϕψ (x, y) = (x, 0), thus showing that (T U, T ϕ) has the subbundle property for ker T f . The same local expression shows that (Tm f )(Tm M ) is closed and splits in Tf (m) M and thus (i) implies (iii). The reverse is proved along the same lines. If f is open (closed) onto its image, then f (U ) ∩ V [resp., int(f (cl(U ))) ∩ V ] can serve as a chart domain for f (M ) with the relative topology induced from N . Thus f (M ) is a submanifold of N .  The relationship between transversality and subimmersivity is given by the following. 3.5.19 Proposition. Let f : M → N be smooth, P a submanifold of N , and assume that f t P . Then there is an open neighborhood W of P in N such that f |f −1 (W ) : f −1 (W ) → W is a subimmersion.

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185

Proof. Let m ∈ f −1 (P ) and write Tm (f −1 (P )) = ker(Tm f ) ⊕ Cm so that Tm f : Cm → Tn P is an isomorphism, where n = f (m). As in the proof of Theorem 3.5.12, this situation can be locally straightened out, so we can assume that M ⊂ E1 × E2 , where P is open in E1 and E1 = ker Tm f ⊕ C for a complementary space C. The map f restricted to C is a local immersion, so projection to C followed by the restriction of f to C writes f as the composition of a submersion followed by an immersion. We leave it to the reader to expand the details of this argument.  In some applications, the closedness of f follows from properness of f ; see the discussion following Theorem 3.5.9. If M or N is finite dimensional, then f being a subimmersion is equivalent to ker T f being a subbundle of T M . Indeed, range(Tm f ) = Tm M/ ker(Tm f ) and thus dim(range(Tm f )) = codim(ker(Tm f )) = constant and f is hence a subimmersion by Proposition 3.5.16(ii). Proposition 3.5.19 is an infinite dimensional version of this. We have already encountered subimmersions in the study of vector bundles. Namely, the condition in Proposition 3.4.18(i) (which insures that for a vector bundle map f over the identity ker f and range f are subbundles) is nothing other than f being a subimmersion. Quotients.

We conclude this section with a study of quotient manifolds.

3.5.20 Definition. An equivalence relation R on a manifold M is called regular if the quotient space M/R carries a manifold structure such that the canonical projection π : M → M/R is a submersion. If R is a regular equivalence relation, then M/R is called the quotient manifold of M by R. Since submersions are open mappings, π and hence the regular equivalence relations R are open. Quotient manifolds are characterized by their effect on mappings. 3.5.21 Proposition.

Let R be a regular equivalent relation on M .

(i) A map f : M/R → N is C r , r ≥ 1 iff f ◦ π : M → N is C r . (ii) Any C r map g : M → N compatible with R, that is, xRy implies g(x) = g(y), defines a unique C r map gˆ : M/R → N such that gˆ ◦ π = g. (iii) The manifold structure of M/R is unique. Proof. (i) If f is C r , then so is f ◦ π by the composite mapping theorem. Conversely, let f ◦ π be C r . Since π is a submersion it can be locally expressed as a projection and thus there exist charts (U, ϕ) at m ∈ M and (V, ψ) at π(m) ∈ M/R such that ϕ(U ) = U1 × U2 ⊂ E1 ⊕ E2 ,

ψ(V ) = U2 ⊂ E2 ,

and πϕψ (x, y) = y.

Hence if (W, χ) is a chart at (f ◦ π)(m) in N satisfying (f ◦ π)(U ) ⊂ W , then fψχ = (f ◦ π)ϕχ |{0} × U2 and thus fψχ is C r . (ii) The mapping gˆ is uniquely determined by gˆ ◦ π = g. It is C r by (i) . (iii) Let (M/R)1 and (M/R)2 be two manifold structures on M/R having π as a submersion. Apply (ii) for (M/R)1 with N = (M/R)2 and g = π to get a unique C ∞ map h : (M/R)1 → (M/R)2 such that h ◦ π = π. Since π is surjective, h = identity. Interchanging the roles of the indices 1 and 2, shows that the identity mapping induces a C ∞ map of (M/R)2 to (M/R)1 . Thus, the identity induces a diffeomorphism.  3.5.22 Corollary. Let M and N be manifolds, R and Q regular equivalence relations on M and N , respectively, and f : M → N a C r map, r ≥ 1, compatible with R and Q; that is, if xRy then f (x)Qf (y). Then f induces a unique C r map ϕ : M/R → N/Q and the diagram

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TM

Tf

T πM

- TN T πN ? - T (N/Q)

? T (M/R) Tφ commutes.

Proof. The map ϕ is uniquely determined by πN ◦ f = ϕ ◦ πM . Since πN ◦ f is C r , ϕ is C r by Proposition 3.5.21. The diagram is obtained by applying the chain rule to πN ◦ f = ϕ ◦ πM .  The manifold M/R might not be Hausdorff. By Proposition 1.4.10 it is Hausdorff iff the graph of R is closed in M × M ; R is open since it is regular. For an example of a non-Hausdorff quotient manifold see Exercise 3.5-8. There is, in fact, a bijective correspondence between surjective submersions and quotient manifolds. More precisely, we have the following. 3.5.23 Proposition. Let f : M → N be a submersion and let R be the equivalence relation defined by f ; that is, xRy iff f (x) = f (y). Then R is regular, M/R is diffeomorphic to f (M ), and f (M ) is open in N . Proof. As f is a submersion, it is an open mapping, so f (M ) is open in N . Moreover, since f is open, so is the equivalence R and thus f induces a homeomorphism of M/R onto f (M ) (see the comments following Definition 1.4.9). Put the differentiable structure on M/R that makes the homeomorphism into a diffeomorphism. Then M/R is a manifold and the projection is clearly a submersion, since f is.  This construction provides a number of examples of quotient manifolds. 3.5.24 Examples. A. The base space of any vector bundle is a quotient manifold. Take the submersion to be the vector bundle projection. B. The circle S 1 is a quotient manifold of R defined by the submersion θ 7→ eiθ ; we can then write S 1 = R/2πZ, including the differentiable structure. C. The Grassmannian Gk (E) is a quotient manifold in the following way. Define the set D by D = { (x1 , . . . , xk ) | xi ∈ E, x1 , . . . , xk are linearly dependent }. Since D is open in the product E × · · · × E (k times), one can define the map π : D → Gk (E) by π(x1 , . . . , xk ) = span{x1 , . . . , xk }. Using the charts described in Example 3.1.8G, one finds that π is a submersion. In particular, the projective spaces RPn and CPn are quotient manifolds. D. The M¨obius band as explained in Example 3.4.10C is a quotient manifold. E. Quotient bundles are quotient manifolds (see Proposition 3.4.17).



We close with an important characterization of regular equivalence relations due to Godement, as presented by Serre [1965]. 3.5.25 Theorem.

An equivalence relation R on a manifold M is regular iff

(i) graph(R) is a submanifold of M × M , and (ii) p1 : graph(R) → M , p1 (x, y) = x is a submersion.

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187

Proof. First assume that R is regular. Since π : M → M/R is a submersion, so is the product π × π : M × M → (M/R) × (M/R) so that graph(R) = (π × π)−1 (∆M/R ), where the set ∆M/R = { ([x], [x]) | [x] ∈ M/R } is the diagonal of M/R×M/R, is a submanifold of M ×M (Theorem 3.5.4). This proves (i). To verify (ii), let (x, y) ∈ graph(R) and vx ∈ Tx M . Since π is a submersion and π(x) = π(y), there exists vy ∈ Ty M such that Ty π(vy ) = Tx π(vx ); that is, (vx , vy ) ∈ T (x, y)(graph(R)) by Theorem 3.5.4. But then T (x, y)p1 (vx , vy ) = vx , showing that p1 is a submersion. To prove the converse, we note that the equivalence relation is open, that is, that π −1 (π(U )) is an open subset of M whenever U is open in M . Indeed π −1 (π(U )) = p1 ((M × U ) ∩ graph(R)) which is open since p1 is an open map being a submersion by (ii). Second, the diffeomorphism s : (x, y) 7→ (y, x) of graph(R) shows that p1 is a submersion iff p2 : graph(R) → M , p2 (x, y) = y is a submersion since p2 = p1 ◦ s. The rest of the proof consists of two major steps: a reduction to a local problem and the solution of the local problem.  S Step 1. If M = i Ui , where Ui are open subsets of M such that Ri = R∩(Ui ×Ui ) is a regular equivalence relation in Ui , then R is regular. Openness of R implies openness of Ri and of Ui∗ = π −1 (π(Ui )). Let us first show that Ri∗ = R ∩ (Ui∗ × Ui∗ ) is regular on Ui∗ . Let πi : Ui → Ui /Ri and πi∗ : Ui∗ → Ui∗ /Ri∗ denote the canonical projections. We prove that the existence of a manifold structure on Ui /Ri and submersivity of πi imply that Ui∗ /Ri∗ has a manifold structure and that πi∗ is a submersion. For this purpose let λi : Ui /Ri → Ui∗ /Ri∗ be the bijective map induced by the inclusion ji : Ui → Ui∗ and endow Ui∗ /Ri∗ with the manifold structure making λi into a diffeomorphism. ∗ ∗ ∗ Thus πi∗ is a submersion iff ρi = λ−1 i ◦ πi : Ui → Ui /Ri is a submersion. Since λi ◦ πi = πi ◦ ji , it follows ∗ that ρi |Ui = πi is a submersion and therefore the composition (ρi |Ui ) ◦ p2 : (Ui × Ui ) ∩ graph(R) → Ui /Ri is a submersion. The relations (ρi |Ui ) ◦ p2 = ρi ◦ p1 show that ρi ◦ p1 is a submersion and since p1 is a surjective submersion this implies that ρi is a submersion (see Exercise 3.5-6(iv)). Thus, in the statement of Step 1, we can assume that all open sets Ui are such that Ui = π −1 (π(Ui )). Let Rij be the equivalence relation induced by R on Ui ∩ Uj . Since Ui ∩ Uj /Rij is open in both Ui /Ri and Uj /Rj , it follows that it has two manifold structures. Since πi and πj are submersions, they will remain submersions when restricted to Ui ∩ Uj . Therefore Rij is regular and by Proposition 3.5.21(iii) the manifold structures on Ui ∩ Uj /Rij induced by the equivalence relations Ui /Ri and Uj /Rj coincide. Therefore there is a unique manifold structure on M/R such that Ui /Ri are open submanifolds; this structure is obtained by collation (see Exercise 3.2-6). The projection π is a submersion since πi = π|Ui is a submersion for all i. Step 2. For each m ∈ M there is an open neighborhood U of m such that R ∩ (U × U ) = RU is regular. The main technical work is contained in the following. 3.5.26 Lemma. For each m ∈ M there is an open neighborhood U of M , a submanifold S of U and a smooth map s : U → S such that [u] ∩ S = {s(u)}; S is called a local slice of R. Let us assume the lemma and use it to prove Step 2. The inclusion of S into U is a right inverse of s and thus s is a submersion. Now define ϕ : S → U/RU by ϕ(u) = [u]. By the lemma, ϕ is a bijective map. Put the manifold structure on U/RU making ϕ into a diffeomorphism. The relation ϕ ◦ s = π|U shows that π|U is submersive and thus RU is regular. Proof of Lemma 3.5.26. In the entire proof, m ∈ M is fixed. Define the space F by F = { v ∈ Tm M | (0, v) ∈ T(m,m) (graph(R)) }, then {0}×F = ker T(m,m) p1 and thus by hypothesis (ii) in the theorem, {0}×F splits in T(m,m) (graph(R)). The latter splits in Tm M × Tm M by hypothesis (i) and thus {0} × F splits in Tm M ×Tm M . Since {0}×F is a closed subspace of {0}×Tm M , it follows that F splits in Tm M (see Exercise 2.1-7). Let G be a closed complement of F in Tm M and choose locally a submanifold P of M, m ∈ P , such

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that Tm P = G. Define the set Q by Q = (M × P ) ∩ graph(R). Since Q = p−1 2 (P ) and p2 : graph(R) → M is a submersion, Q is a submanifold of graph(R). We claim that T(m,m) p1 : T(m,m) Q → Tm M is an isomorphism. Since T(m,m) Q = (T(m,m) p2 )−1 (Tm P ), it follows that ker(T(m,m) p1 |T(m,m) Q) = ker T(m,m) p1 ∩ (T(m,m) p2 )−1 (Tm P ) = ({0} × F ) ∩ (Tm M × G) = {0} × (F ∩ G) = {(0, 0)}, that is, T(m,m) p1 |T(m,m) Q is injective. Now let u ∈ Tm M and choose v ∈ Tm M such that (u, v) ∈ T(m,m) (graph(R)). If v = v1 + v2 , v1 ∈ F, v2 ∈ G, then (u, v2 ) = (u, v) − (0, v1 ) ∈ T(m,m) (graph(R)) + ({0} × F) ⊂ T(m,m) (graph(R)) and T(m,m) p2 (u, v2 ) = v2 ∈ Tm P , that is, (u, v2 ) ∈ (T(m,m) p2 )−1 (Tm P ) = T(m,m) Q. Then T(m,m) p1 (u, v2 ) = u and hence T(m,m) p1 |T(m,m) Q is onto. Thus p1 : Q → M is a local diffeomorphism at (m, m), that is, there are open neighborhoods U1 and U2 of m, U1 ⊂ U2 such that p1 : Q ∩ (U1 × U2 ) → U1 is a diffeomorphism. Let σ be the inverse of p1 on U1 . Since σ is of the form σ(x) = (x, s(x)), this defines a smooth map s : U1 → P . Note that if x ∈ U1 ∩ P , then (x, x) and (x, s(x)) are two points in Q ∩ (U1 × U2 ) with the same image in U1 and hence are equal. This shows that s(x) = x for x ∈ U1 ∩ P . Set U = { x ∈ U1 | s(x) ∈ U1 ∩ P } and let S = U ∩ P . Since s is smooth and U1 ∩ P is open in P it follows that U is open in U1 hence in M . Also, m ∈ U since m ∈ U1 ∩ P and so m = s(m). Let us show that s(U ) ⊂ S, that is, that if x ∈ U , then s(x) ∈ U and s(x) ∈ P . The last relation is obvious from the definition of U . To show that s(x) ∈ U is equivalent to proving that s(x) ∈ U1 , which is clear, and that s(s(x)) ∈ U1 ∩ P . However, since s(s(x)) = s(x), because s(x) ∈ P , it follows from x ∈ U that s(x) ∈ U1 ∩ P . Thus we have found an open neighborhood U of m, a submanifold S of U , and a smooth map s : U → S which is the identity in U ∩ S. Finally, we show that s(x) is the only element of S equivalent to x ∈ U . But this is clear since there is exactly one point in (U × S) ∩ graph(R), namely (x, s(x)) mapped by p1 into x, since p1 |(U × S) ∩ graph(R) is a diffeomorphism m.  The above proof shows that in condition (ii) of the theorem, p1 can be replaced by p2 . Also recall that M/R is Hausdorff iff R is closed.

Supplement 3.5A Lagrange Multipliers Let M be a smooth manifold and i : N → M a submanifold of M , i denoting the inclusion mapping. If f : M → R, we want to determine necessary and sufficient conditions for n ∈ N to be a critical point of f |N , the restriction of f to N . Since f |N = f ◦ i, the chain rule gives Tn (f |N ) = Tn f ◦ Tn i; thus n ∈ N is a critical point of f |N iff Tn f |Tn N = 0. This condition takes a simple form if N happens to be the inverse image of a point under submersion. 3.5.27 Theorem (Lagrange Multiplier Theorem). Let g : M → P be a smooth map and let p ∈ P be a regular value of g. Let N = g −1 (p) and let f : M → R be C r , r ≥ 1. A point n ∈ N is a critical point of f |N if there exists λ ∈ Tp∗ P , called a Lagrange multiplier , such that Tn f = λ ◦ Tn g.

3.5 Submersions, Immersions, and Transversality

Proof.

189

First assume such a λ exists. Since Tn N = ker Tn g, (λ ◦ Tn g ◦ Tn i)(vn ) = λ(Tn g(vn )) = 0

for all vn ∈ Tn N ;

that is, 0 = (λ ◦ Tn g)|Tn N = Tn f |Tn N . Conversely, assume Tn f |Tn N = 0. By the local normal form for submersions, there is a chart (U, ϕ) at n, ϕ : U → U1 × V1 ⊂ E × F such that ϕ(U ∩ N ) = {0} × V1 satisfying ϕ(n) = (0, 0), and a chart (V, ψ) at p, ψ : V → U1 ⊂ E where g(U ) ⊂ V , ψ(p) = 0, and such that gϕψ (x, y) = (ψ ◦ g ◦ ϕ−1 )(x, y) = x for all (x, y) ∈ U1 × V1 . If fϕ = f ◦ ϕ−1 : U1 × V1 → R, we have for all v ∈ F, D2 fϕ (0, 0) · v = 0 since Tn f |Tn N = 0. Thus, letting µ = D1 fϕ (0, 0) ∈ E∗ , u ∈ E and v ∈ F, we get Dfϕ (0, 0) · (u, v) = µ(u) = (µ ◦ Dgϕψ )(0, 0) · (u, v); that is, Dfϕ (0, 0) = (µ ◦ Dgϕψ )(0, 0). To pull this local calculation back to M and P , let λ = µ ◦ Tp ψ ∈ Tp∗ P , so composing the foregoing relation with Tn ϕ on the right we get Tn f = λ ◦ Tn g.  3.5.28 Corollary. Let g : M → P be transversal to the submanifold W of P , N = g −1 (W ), and let f : M → R be C r , r ≥ 1. Let Eg(n) be a closed complement to Tg(n) W in Tg(n) P so Tg(n) P = Tg(n) W ⊕ Eg(n) ∗ and let π : Tg(n) P → Eg(n) be the projection. A point n ∈ N is a critical point of f |N iff there exists λ ∈ Eg(n) called a Lagrange multiplier such that Tn f = λ ◦ π ◦ Tn g. Proof. By Theorem 3.5.12, there is a chart (U, ϕ) at n, with ϕ(U ) = U1 ×U2 ⊂ E1 ×E2 , ϕ(U ∩N ) = {0}× U2 , and ϕ(n) = (0, 0), and a chart (V, ψ) at g(n) satisfying ψ(V ) = U1 × V1 ⊂ F1 × F, ψ(V ∩ W ) = {0} × V1 , ψ(g(n)) = (0, 0), and g(U ) ⊂ V , such that the local representative satisfies gϕψ (x, y) = (ψ ◦ g ◦ ϕ−1 )(x, y) = (x, η(x, y)) for all (x, y) ∈ U1 ×V1 . Let ρ : E1 ×F → E1 be the canonical projection. By the Lagrange multiplier theorem applied to the composition ρ ◦ gϕψ : U1 × U2 → U1 , (0, 0) ∈ U1 × U2 is a critical point of f |{0} × U2 iff there is a point µ ∈ E∗1 such that Dfϕ (0, 0) = µ ◦ ρ ◦ Dgϕψ (0, 0). Composing this relation on the right with Tn ϕ and letting λ = µ ◦ Tg(n) ψ and π = (Tg(n) ψ)−1 |Eg(n) ◦ ρ ◦ Tg(n) ψ : Tg(n) P → Eg(n) , we get the required identity Tn f = λ ◦ π ◦ Tn g.



If P is a Banach space F, then Theorem 3.5.27 can be formulated in the following way. 3.5.29 Corollary. Let F be a Banach space, g : M → F be a smooth map having 0 as a regular value, N = g −1 (0), and f : M → R be C r , r ≥ 1. The point n ∈ N is a critical point of f |N iff there exists λ ∈ F∗ , called a Lagrange multiplier , such that n is a critical point of f − λ ◦ g. Remarks. 1. λ depends not just on f |N but also on how f is extended off N . 2. This form of the Lagrange multiplier theorem is extensively used in the calculus of variations to study critical points of functions with constraints; cf. Caratheodory [1965]. 3. We leave it to the reader to generalize Corollary 3.5.28 in the same spirit.

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4. There are generalizations to f : M → Rk , which we invite the reader to formulate. The name Lagrange multiplier is commonly used in conjunction with the previous corollary in Euclidean spaces. Let U be an open set in Rn , F = Rp , g = (g 1 , . . . , g p ) : U → Rp a submersion and f : U → R smooth. Then x ∈ N = g −1 (0) is a critical point of f |N , iff there exists λ=

p X

λi ei ∈ (Rp )∗ ,

i=1

where e1 , . . . , ep is the standard dual basis in Rp , such that n is a critical point of f −λ◦g =f −

p X

λi g i .

i=1

In calculus, the real numbers λi are referred to as Lagrange multipliers. Thus, to find a critical point x = (x1 , . . . , xm ) ∈ N ⊂ Rm of f |N one solves the system of m + p equations p X ∂g i ∂f (x) − λi j (x) = 0, j ∂x ∂x i=1

g i (x) = 0,

j = 1, . . . , m i = 1, . . . , p

for the m + p unknowns x1 , . . . , xm , λ1 , . . . , λp . For example, let N = S 2 ⊂ R3 and f : R3 → R; f (x, y, z) = z. Then f |S 2 is the height function on the sphere and we would expect (0, 0, ±1) to be the only critical points of f |S 2 ; note that f itself has no critical points. The method of Lagrange multipliers, with g(x, y, z) = x2 + y 2 + z 2 − 1, gives 0 − 2xλ = 0,

0 − 2yλ = 0,

1 − 2zλ = 0,

and x2 + y 2 + z 2 = 1.

The only solutions are λ = ±1/2, x = 0, y = 0, z = ±1, and indeed these correspond to the maximum and minimum points for f on S 2 . See an elementary text such as Marsden and Tromba [1996] for additional examples. For more advanced applications, see Luenberger [1969]. The reader will recall from advanced calculus that maximum and minimum tests for a critical point can be given in terms of the Hessian, that is, matrix of second derivatives. For constrained problems there is a similar test involving bordered Hessians. Bordered Hessians are simply the Hessians of h = f − λg + c(λ − λ0 )2 in (x, λ)-space. Then the Hessian test for maxima and minima apply; a maximum or minimum of h clearly implies the same for f on a level set of g. See Marsden and Tromba [1996] (pp. 224–30) for an elementary treatment and applications.

Exercises  3.5-1. (i) Show that the set SL(n, R) of elements of L(Rn , Rn ) with determinant 1 is a closed submanifold of dimension n2 − 1; SL(n, R) is called the special linear group. Generalize to the complex case. (ii) Show that O(n) has two connected components. The component of the identity SO(n) is called the special orthogonal group. (iii) Let U(n) = { U ∈ L(Cn , Cn ) | U U ∗ = Identity } be the unitary group. Show that U(n) is a non-compact submanifold of dimension n2 of L(Cn , Cn ) and O(2n). (iv) Show that the special unitary group SU(n) = U(n)∩SL(n, C) is a compact n2 −1 dimensional manifold.

3.5 Submersions, Immersions, and Transversality

191

(v) Define  J=

0 −I

 I ∈ L(R2n , R2n ), 0

where I is the identity of Rn . Show that the space Sp(2n, R) defined by Sp(2n, R) = { Q ∈ L(R2n , R2n ) | QJQT = J } is a compact submanifold of dimension 2n2 + n; it is called the symplectic group.  3.5-2. Define USt(n, n; k) and Sp St(2n, 2n; 2k) analogous to the definition of OSt(n, n; k) in Example 3.5.5D. Show that they are compact manifolds and compute their dimensions.  3.5-3.

(i) Let P ⊂ O(3) be defined by P = { Q ∈ O(3) | det Q = +1, Q = QT }\{I}.

Show that P is a two-dimensional compact submanifold of O(3). (ii) Define f : RP2 → O(3), f (`) = the rotation through π about the line `. Show that f is a diffeomorphism of RP2 onto P.  3.5-4. (i) If N is a submanifold of dimension n in an m-manifold M , show that for each x ∈ N there is an open neighborhood U ⊂ M with x ∈ U and a submersion f : U ⊂ M → Rm−n , such that N ∩ U = f −1 (0). (ii) Show that RP1 is a submanifold of RP2 , which is not the level set of any submersion of RP2 into RP1 ; in fact, there are no such submersions. Hint: RP1 is one-sided in RP2 .  3.5-5. (i) Show that if f : M → N is a subimmersion, g : N → P an immersion and h : Z → M a submersion, then g ◦ f ◦ h is a subimmersion. (ii) Show that if fi : Mi → Ni , i = 1, 2 are immersions (submersions, subimmersions), then so is f1 × f2 : M1 × M2 → N 1 × N 2 . (iii) Show that the composition of two immersions (submersions) is again an immersion (submersion). Show that this fails for subimmersions. (iv) Let f : M → N and g : N → P be C r , r ≥ 1. If g ◦ f is an immersion, show that f is an immersion. If g ◦ f is a submersion and if f is onto, show that g is a submersion. (v) Show that if f is an immersion (resp., embedding, submersion, subimmersion) then so is T f .  3.5-6. (i) Let M be a manifold, R a regular equivalence relation and S another equivalence relation implied by R; that is, graph R ⊂ graph S. Denote by S/R the equivalence relation induced by S on M/R. Show that S is regular iff S/R is and in this case establish a diffeomorphism (M/R)/(S/R) → M/S. (ii) Let Mi , i = 1, 2 be manifolds and Ri be regular equivalences on Mi . Denote by R the equivalence on M1 × M2 defined by R1 × R2 . Show that M/R is diffeomorphic to (M1 /R1 ) × (M2 /R2 ).  3.5-7 (The line with two origins). Let M be the quotient topological space obtained by starting with (R × {0}) ∪ (R × {1}) and identifying (t, 0) with (t, 1) for t 6= 0. Show that this is a one-dimensional non-Hausdorff manifold. Find an immersion R → M .  3.5-8. Let f : M → N be C ∞ and denote by h : T M → f ∗ (T N ) the vector bundle map over the identity uniquely defined by the pull-back. Prove the following:

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3. Manifolds and Vector Bundles h

(i) f is an immersion iff 0 → T M → f ∗ (T N ) is fiber split exact; h

(ii) f is a submersion iff T M → f ∗ (T N ) → 0 is fiber split exact; (iii) f is a subimmersion iff ker(h) and range(h) are subbundles.  3.5-9. Let A be a real nonsingular symmetric n × n matrix and c a nonzero real number. Show that the quadratic surface { x ∈ Rn | hAx, xi = c } is an (n − 1)-submanifold of Rn .  3.5-10 (Steiner’s Roman Surface).

Let f : S 2 → R4 be defined by

f (x, y, z) = (yz, xz, xy, x2 + 2y 2 + 3z 2 ). (i) Show that f (p) = f (q) if and only if p = ±q. (ii) Show that f induces an immersion f 0 : RP2 → R4 . (iii) Let g : RP2 → R3 be the first three components of f 0 . Show that g is a “topological” immersion and try to draw the surface g(RP2 ) (see Spivak [1979] for the solution).  3.5-11 (Covering maps). Let f : M → N be smooth and M compact, dim(M ) = dim(N ) < ∞. If n is a regular value of f , show that f −1 (n) is a finite set {m1 , . . . , mk } and that there exists an open neighborhood V of n in N and disjoint open neighborhoods U1 , . . . , Uk of m1 , . . . , mk such that f −1 (V ) = U1 ∪ · · · ∪ Uk , and f |Uk : Ui → V , i = 1, . . . , k are all diffeomorphisms. Show k is constant if M is connected and f is a submersion.  3.5-12. Let f : M → N and g : N → P be smooth maps, such that g t V where V is a submanifold of P . Show that f t g −1 (V ) iff g ◦ f t V .  3.5-13. Show that an injective immersion f : M → N is an embedding iff f (M ) is a closed submanifold of an open submanifold of N . Show that if f : M → N is an embedding, f is a diffeomorphism of M onto f (M ). Hint: See Exercise 2.5-12.  3.5-14.

Show that the map p : St(n, n; k) = Gk (Rn ) defined by p(A) = range A is a surjective submersion.

 3.5-15.

Show that f : RPn × RPm → RPmn+m+n given by (x, y) 7→ [x0 y0 , x0 y1 , . . . , xi yj , . . . , xn ym ]

is an embedding. (This embedding is used in algebraic geometry to define the product of quasiprojective varieties; it is called the Segre embedding .)  3.5-16.

Show that { (x, y) ∈ RPn × RPm | n ≤ m,

X

xi yi = 0 }

i=0,...,n

is an (m + n − 1)-manifold. It is usually called a Milnor manifold .  3.5-17 (Fiber product of manifolds). Let f : M → P and g : N → P be C ∞ mappings such that (f, g) : M × N → P × P is transversal to the diagonal of P × P . Show that the set defined by M ×P N = { (m, n) ∈ M × N | f (m) = g(n) } is a submanifold of M × N . If M and N are finite dimensional, show that codim(M ×P N ) = dim P .  3.5-18. (i) Let H be a Hilbert space and GL(H) the group of all isomorphisms A : H → H that are continuous. As we saw earlier, GL(H) is open in L(H, H) and multiplication and inversion are C ∞ maps, so GL(H) is a Lie group. Show that O(H) ⊂ GL(H) defined by O(H) = { A ∈ GL(H) | hAx, Ayi = hx, yi for all x, y ∈ H } is a smooth submanifold and hence also a Lie group.

3.5 Submersions, Immersions, and Transversality

193

(ii) Show that the tangent space at the identity of GL(H) (the Lie algebra) consists of all bounded skew adjoint operators, as follows. Let S(H) = { A ∈ L(H, H) | A∗ = A }, where A∗ is the adjoint of A. Define f : GL(H) → S(H), by f (A) = A∗ A. Show f is C ∞ , f −1 (I) = O(H), and Df (A) · B = B ∗ A + A∗ B. Show that f is a submersion, and ker Df (A) = { B ∈ L(H, H) | B ∗ A + A∗ B = 0 }, which splits; a complement is the space { T ∈ L(H, H) | T ∗ A = A∗ T } since any U splits as U=

1 1 (U − A∗−1 U ∗ A) + (U + A∗−1 U ∗ A). 2 2

(i) If f : M → N is a smooth map of finite-dimensional manifolds and m ∈ M , show that there  3.5-19. is an open neighborhood U of m such that rank(Tx f ) ≥ rank(Tm f ) for all x ∈ U . Hint: Use the local expression of Tm f as a Jacobian matrix. (ii) Let M be a finite-dimensional connected manifold and f : M → M a smooth map satisfying f ◦ f = f . Show that f (M ) is a closed connected submanifold of M . What is its dimension? Hint: Show f is a closed map. For m ∈ f (M ) show that range(Tm f ) = ker(Identity −Tm f ) and thus rank(Tm f ) + rank(Identity −Tm f ) = dim M. Both ranks can only increase in a neighborhood of m by (i), so rank(Tm f ) is locally constant on f (M ). Thus there is a neighborhood U of f (M ) such that the rank of f on U is bigger than or equal to the rank of f on f (M ). Use rank(AB) ≤ rank A and the fact that f ◦ f = f to show that the rank of f on U is smaller than or equal to the rank of f in f (M ). Therefore rank of f on U is constant. Apply Theorem 3.5.18(iii).  3.5-20. (i) Let α, β : E → R be continuous linear maps on a Banach space E such that ker α = ker β. Show that α and β are proportional. Hint: Split E = ker α ⊕ R. (ii) Let f, g : M → R be smooth functions with 0 a regular value of both f and g and N = f −1 (0) = g −1 (0). Show that for all x ∈ N , df (x) = λ(x) dg(x) for a smooth function λ : N → R.  3.5-21. Let f : M → N be a smooth map, P ⊂ N a submanifold, and assume f t P . Use Definition 3.5.10 and Exercise 3.4-10 to show the vector bundle isomorphism ν(f −1 (P )) ∼ = f ∗ (ν(P )). −1 Hint: Look at T f |f (P ) and compute the kernel of the induced map T M |f −1 (P ) → ν(P ). Obtain a vector bundle map ν(f −1 (P )) → ν(P ) which is an isomorphism on each fiber. Then invoke the universal property of the pull–back of vector bundles; see Exercise 3.4-13.  3.5-22. (i) Recall (Exercise 3.2-1) that S n with antipodal points identified is diffeomorphic to RPn . Conclude that any closed hemisphere of S n with antipodal points on the great circle identified is also diffeomorphic to RPn . (ii) Let B n be the closed unit ball in Rn . Map B n to the upper hemisphere of S n by mapping x 7→ (x, (1 − kxk2 )1/2 ). Show that this map is a diffeomorphism of an open neighborhood of B n to an open n neighborhood of the upper hemisphere S+ = { x ∈ S n | xn+1 ≥ 0 }, mapping B n homeomorphically n to S+ and homeomorphically to the great circle { x ∈ S n | xn+1 = 0 }. Use (i) to show that B n with antipodal points on the boundary identified is diffeomorphic to RPn .

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3. Manifolds and Vector Bundles

(iii) Show that SO(3) is diffeomorphic to RP3 ; the diffeomorphism is induced by the map sending the closed unit ball B 3 in R3 to SO(3) via (x, y, z) 7→ (the rotation about (x, y, z) by the right hand rule in the plane perpendicular to (x, y, z) through the angle π(x2 + y 2 + z 2 )1/2 ).  3.5-23.

(i) Show that S n × R embeds in Rn+1 .

Hint: The image is a “fat” sphere.

(ii) Describe explicitly in terms of trigonometric functions the embedding of T2 into R3 . (iii) Show that S a(1) × · · · × S a(k) , where a(1) + · · · a(k) = n embeds in Rn+1 . Hint: Show that its product with R embeds in Rn+1 by (i).  3.5-24. Let f : M → M be an involution without fixed points, that is, f ◦ f = identity and f (m) 6= m for all m. Let R be the equivalence relation determined by f , that is, m1 Rm2 iff f (m1 ) = f (m2 ). (i) Show R is a regular equivalence relation. (ii) Show that the differentiable structure of M/R is uniquely determined by the property: the projection π : M → M/R is a local diffeomorphism.  3.5-25 (Connected sum of manifolds). Let M and N to be two Hausdorff manifolds modeled on the same Banach space E. Let m ∈ M , n ∈ N and let (U0 , ϕ0 ) be a chart at m and let (V0 , ψ0 ) be a chart at n such that ϕ0 (m) = ψ0 (n) = 0 and ϕ0 (U ), ψ0 (V ) contain the closed unit ball in E. Thus, if B denotes the open unit ball in E, ϕ0 (U )\B and ψ0 (V )\B are nonempty. If A and B are atlases of M and N respectively, let Am , Bn be the induced atlases on M \{m} and N \{n} respectively. Define     x x σ : B\{0} → B\{0} by σ kxk, = 1 − kxk, kxk kxk and observe that σ 2 = identity. Let W be the disjoint union of M \{m} with N \{n} and define an equivalence relation R in W by v1 Rv2 iff (w1 = w2 ) or (w1 ∈ M \{m}, ϕ0 (w1 ) ∈ B\{0} and w2 ∈ N \{n}, ψ0 (w2 ) ∈ B\{0} and ϕ0 (w1 ) = (σ ◦ ψ0 )(w2 )) or (same condition with w1 and w2 interchanged). (i) Show that R is an equivalence relation on W . (ii) Show R is regular. (iii) If π : W → W/R denotes the projection, show that for any open set O in the atlas of W , π : O 7→ π(O) is a diffeomorphism. (iv) Show that { (π(U ), ϕ ◦ (π|U )−1 ), (π(V ), ψ ◦ (π|V )−1 ) | (U, ϕ) ∈ Am , (V, ψ) ∈ Bn } is an atlas defining the differentiable structure of W/R. (v) W/R is denoted by M #N . Draw T2 #T2 and identify T2 #RP2 and RP2 #RP2 . (vi) Prove that M #(N #P ) ≈ (M #N )#P,

M #N ≈ N #M,

M #SE ≈ M,

where M , N , and P are all modeled on E, SE is the unit sphere in E, and ≈ denotes “diffeomorphic.” (vii) Compute Rn #Rn # · · · #Rn (k times) for all positive integers n and k and show that it embeds in Rn .

3.6 The Sard and Smale Theorems

 3.5-26.

195

(i) Let a > 0 and define  χa : R → R

by χa (x) = exp

−x2 a2 − x2

 ,

if x ∈ ]−a, a[ and χa (x) = 0 if x ∈ R\]−a, a[. Show that this is a C ∞ function and satisfies the inequalities 0 ≤ χa (x) ≤ 1, |χ0a (x)| < 1 for all x ∈ R, and χa (0) = 1. (ii) Fix a > 0 and λ ∈ E∗ , where E is a Banach space whose norm is of class C r away from the origin and r ≥ 1. Write E = ker λ ⊕ R; this is always possible since any closed finite codimensional space splits (see §2.2). Define, for any t ∈ R, fλ,a,t : E → E by fλ,a,t (u, x) = (u, x + tχa (kuk)) where u ∈ ker λ and x ∈ R. Show that fλ,a,t satisfies fλ,a,t (0, 0) = (0, 1) and fλ,a,t |(E\ cl(Ba (0))) = identity. Hint: Show that fλ,a,t is a bijective local diffeomorphism. (iii) Let M be a C r Hausdorff manifold modeled on a Banach space E whose norm is C r on E\{0}, r ≥ 1. Assume dim M ≥ 2. Let C be a closed set in M and assume that M \C is connected. Let {p1 , . . . , pk }, {q1 , . . . , qk } be two finite subsets of M \C. Show that there exists a C r diffeomorphism ϕ : M → M such that ϕ(pi ) = qi , i = 1, . . . , k and ϕ|C = identity. Show that if k = 1, the result holds even if dim M = 1. Hint: For k = 1, define an equivalence relation on M \C : m ∼ n iff there is a diffeomorphism ψ : M → M homotopic to the identity such that ϕ(m) = n and ψ|C = identity. Show that the equivalence classes are open in M \C in the following way. Let ϕ : U → E be a chart at m, ϕ(m) = 0, U ⊂ M \C, and let n ∈ U , n 6= m. Use the Hahn–Banach theorem to show that there is λ ∈ E∗ such that ϕ can be modified to satisfy ϕ(m) = (0, 0), ϕ(n) = (0, 1), where E = ker λ ⊕ R. Use (ii) to find a diffeomorphism h : U → U homotopic to the identity on U , satisfying h(m) = n and h|(U \A) = identity, where A is a closed neighborhood of n. Then f : M → M which equals h on U and the identity on M \U establishes m ∼ n. For general k proceed by induction, using the connectedness of M \C\{q1 , . . . , qk−1 } and finding by the case k = 1 a diffeomorphism g homotopic to the identity on M sending h(pk ) to qk and keeping C ∪ {q1 , . . . , qk } fixed; h : M → M is the diffeomorphism given by induction which keeps C fixed and sends pi to qi for r = 1, . . . , k − 1. Then f = g ◦ h is the desired diffeomorphism.

3.6

The Sard and Smale Theorems

This section is devoted to the classical Sard theorem and its infinite-dimensional generalization due to Smale [1965]. We first develop a few properties of sets of measure zero in Rn . Sets of Measure Zero. A subset A ⊂ Rm is said to have measure zero if, for every ε > 0, there is a countable covering of A by open sets Ui , such that the sum of the volumes of Ui is less than ε. Clearly a countable union of sets of measure zero has measure zero. 3.6.1 Lemma. Let U ⊂ Rm be open and A ⊂ U be of measure zero. If f : U → Rm is a C 1 map, then f (A) has measure zero. Proof. Let A be contained in a countable union of relatively compact sets Cn . If we show that f (A ∩ Cn ) has measure zero, then f (A) has measure zero since it will be a countable union of sets of measure zero. But Cn is relatively compact and thus there exists M > 0 such that kDf (x)k ≤ M for all x ∈ C√ n . By the mean value theorem, the image of a cube of edge length e is contained in a cube of edge length e mM .  3.6.2 Lemma (Fubini Lemma). Let A be a countable union of compact sets in Rn , fix an integer r satisfying 1 ≤ r ≤ n − 1 and assume that Ac = A ∩ ({c} × Rn−r ) has measure zero in Rn−r for all c ∈ Rr . Then A has measure zero.

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Proof. By induction we reduce to the case r = n − 1. It is enough to work with one element of the union, so we may assume A itself is compact and hence there exists and interval [a, b] such that A ⊂ [a, b] × Rn−1 . Since Ac is compact and has measure zero for each c ∈ [a, b], there is a finite number of closed cubes Kc,1 , . . . , Kc,N (c) in Rn−1 the sum of whose volumes is less than ε and such that {c} × Kc,i cover Ac , i = 1, . . . , N (c). Find a closed interval Ic with c in its interior such that Ic × Kc,i ⊂ Ac × Rn−1 . Thus the family { Ic × Kc,i | i = 1, . . . , N (c), c ∈ [a, b] } covers A ∩ ([a, b] × Rn−1 ) = A. Since { int(Ic ) | c ∈ [a, b] } covers [a, b], we can choose a finite subcovering Ic(1) , . . . , Ic(M ) . Now find another covering Jc(1) , . . . , Jc(K) such that each Jc(i) is contained in some Ic(j) and such that the sum of the lengths of all Jc(i) is less than 2(b − a). Consequently { Jc(j) × Kc(j),i | j = 1, . . . , K, i = 1, . . . , Nc(j) } cover A and the sum of their volumes is less than 2(b − a)ε.  Sard Theorem. Let us recall the following notations from §3.5. If M and N are C 1 manifolds and f : M → N is a C 1 map, a point x ∈ M is a regular point of f if Tx f is surjective, otherwise x is a critical point of f . If C ⊂ M is the set of critical points of f , then f (C) ⊂ N is the set of critical values of f and N \f (C) is the set of regular values of f , which is denoted by Rf or R(f ). In addition, for A ⊂ M we define Rf |A by Rf |A = N \f (A ∩ C). In particular, if U ⊂ M is open, Rf |U = R(f |U ). 3.6.3 Theorem (Sard’s Theorem in Rn ). Let U ⊂ Rm be open and f : U → Rn be of class C k , where k > max(0, m − n). Then the set of critical values of f has measure zero in Rn . Note that if m ≤ n, then f is only required to be at least C 1 . Proof.

(Complete only for k = ∞) Denote by C = { x ∈ U | rank Df (x) < n }

the set of critical points of f . We shall show that f (C) has measure zero in Rn . If m = 0, then Rm is one point and the theorem is trivially true. Suppose inductively the theorem holds for m − 1. Let Ci = { x ∈ U | Dj f (x) = 0 for j = 1, . . . , i }, and write C as the following union of disjoint sets: C = (C\C1 ) ∪ (C1 \C2 ) ∪ · · · ∪ (Ck−1 \Ck ) ∪ Ck . The proof that f (C) has measure zero is divided in three steps. 1. f (Ck ) has measure zero. 2. f (C\C1 ) has measure zero. 3. f (Cs \Cs+1 ) has measure zero, where 1 ≤ s ≤ k − 1. Proof of Step 1. Since k ≥ 1, kn ≥ n + k − 1. But k ≥ m − n + 1, so that kn ≥ m. Let K ⊂ U be a closed cube with edges parallel to the coordinate axes. We will show that f (Ck ∩ K) has measure zero. Since Ck can be covered by countably many such cubes, this will prove that f (Ck ) has measure zero. By Taylor’s theorem, the compactness of K, and the definition of Ck , we have f (y) = f (x) + R(x, y)

where kR(x, y)k ≤ M ky − xkk+1

(3.6.1)

for x ∈ Ck ∩ K and y ∈ K. Here M is a constant depending only on Dk f and K. Let e be the edge 0 length of K. Choose an integer `, subdivide K into `m cubes with edge e/`, and choose √ any cube K of this subdivision which intersects Ck . For x ∈ Ck ∩ K 0 and y ∈ K 0 , we have kx − yk ≤ m(e/`). By equation (3.6.1), f (K 0 ) ⊂ L where L is the cube of edge N `k−1 with center f (x); N = 2M ((m)1/2 `)k+1 . The volume of L is N n `−n(k+1) . There are at most `m such cubes; hence, f (Ck ∩ K) is contained in a union of cubes whose total volume V satisfies V ≤ N n `m−n(k+1) . Since m ≤ kn, m − n(k + 1) < 0, so V → 0 as ` → ∞, and thus f (Ck ∩ K) has measure zero.

3.6 The Sard and Smale Theorems

Proof of Step 2.

197

Write C\C1 = { x ∈ U | 1 ≤ rank Df (x) < n } = K1 ∪ · · · ∪ Kn−1 ,

where Kq = { x ∈ U | rank Df (x) = q } and it suffices to show that f (Kq ) has measure zero for q = 1, . . . , n − 1. Since Kq is empty for q > m, we may assume q ≤ m. As before it will suffice to show that each point Kq has a neighborhood V such that f (V ∩ Kq ) has measure zero. Choose x0 ∈ Kq . By the local representation theorem 2.5.14, we may assume that x0 has a neighborhood V = V1 × V2 , where V1 ⊂ Rq and V2 ⊂ Rm−q are open balls, such that for t ∈ V1 and x ∈ V2 , f (t, x) = (t, η(t, x)). Hence η : V1 × V2 → Rn−q is a C k map. For t ∈ V1 define ηt : V2 → Rn−q by ηt (x) = η(t, x) for x ∈ V2 . Then for every t ∈ V1 , Kq ∩ ({t} × V2 ) = {t} × { x ∈ V2 | Dηt (x) = 0 }. This is because, for (t, x) ∈ V1 × V2 , Df (t, x) is given by the matrix   Iq 0 Df (t, x) = . ∗ Dηt (x) Hence rank Df (t, x) = q iff Dηt (x) = 0. Now ηt is C k and k ≥ m − n = (m − q) − (n − q). Since q ≥ 1, by induction we find that the critical values of ηt , and in particular ηt ({ x ∈ V2 | Dηt (x) = 0 }), has measure zero for each t ∈ V2 . By Fubini’s lemma, f (Kq ∩ V ) has measure zero. Since Kq is covered by countably many such V , this shows that f (Kq ) has measure zero. Proof of Step 3. To show f (Cs \Cs+1 ) has measure zero, it suffices to show that every x ∈ Cs \Cs+1 has a neighborhood V such that f (Cs ∩ V ) has measure zero; then since Cs \Cs+1 is covered by countably many such neighborhoods V , it follows that f (Cs \Cs+1 ) has measure zero. Choose x0 ∈ Cs \Cs+1 . All the partial derivatives of f at x0 of order less than or equal to s are zero, but some partial derivative of order s + 1 is not zero. Hence we may assume that D1 w(x0 ) 6= 0 and w(x0 ) = 0, where D1 is the partial derivative with respect to x1 and that w has the form w(x) = Di(1) · · · Di(s) f (x). Define h : U → Rm by h(x) = (w(x), x2 , . . . , xm ), m

where x = (x1 , x2 , . . . , xm ) ∈ U ⊂ R . Clearly h is C k−s and Dh(x0 ) is nonsingular; hence there is an open neighborhood V of x0 and an open set W ⊂ Rm such that h : V → W is a C k−s diffeomorphism. Let A = Cs ∩ V , A0 = h(A) and g = h−1 . We would like to consider the function f ◦ g and then arrange things such that we can apply the inductive hypothesis to it. If k = ∞, there is no trouble. But if k < ∞, then f ◦ g is only C k−s and the inductive hypothesis would not apply anymore. However, all we are really interested in is that some C k function F : W → Rn exists such that F (x) = (f ◦ g)(x) for all x ∈ A0 and DF (x) = 0 for all x ∈ A0 . The existence of such a function is guaranteed by the Kneser–Glaeser rough composition theorem (Abraham and Robbin [1967]). For k = ∞, we take F = f ◦ g. In any case, define the open set W0 ⊂ Rm−1 by W0 = { (x2 , . . . , xm ) ∈ Rm−1 | (0, x2 , . . . , xn ) ∈ W } and F0 : W0 → Rm by F0 (x2 , . . . , xm ) = F (0, x2 , . . . , xm ) Let S = { (x2 , . . . , xm ) ∈ W0 | DF0 (x2 , . . . , xm ) = 0 }.

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3. Manifolds and Vector Bundles

By the induction hypothesis, F0 (S) has measure zero. But A0 = h(Cs ∩ V ) ⊂ 0 × S since for x ∈ A0 , DF (x) = 0 and since for x ∈ Cs ∩ V , h(x) = (w(x), x2 , . . . , xm ) = (0, x2 , . . . , xm ) because w is an sth derivative of f . Hence f (Cs ∩ V ) = F (h(Cs ∩ V )) ⊂ F (0 × S) = F0 (S), and so f (Cs ∩ V ) has measure zero. As Cs \Cs+1 is covered by countably many such V , the sets f (Cs \Cs+1 ) have measure zero (s = 1, . . . , k − 1).  The smoothness assumption k ≥ 1 + max(0, m − n) cannot be weakened as the following counterexample shows. 3.6.4 Example (Devil’s Staircase Phenomenon). The Cantor set C is defined by the following construction. Remove the open interval ]−1/3, 2/3[ from the closed interval [0, 1]. Then remove the middle thirds ]1/9, 2/9[ and ]7/9, 8/9[ from the closed intervals [0, 1/3] and [2/3, 1] respectively and continue this process of removing the middle third of each remaining closed interval indefinitely. The set C is the remaining set. Since we have removed Pa (countable) union of open intervals, C is closed . The total length of the removed intervals equals (1/3) n≥0 (2/3)n = 1 and thus C has measure zero in [0, 1]. On the other hand each point of C is approached arbitrarily closely by a sequence of endpoints of the intervals removed, that is, each point of C is an accumulation point of [0, 1]\C. Each open subinterval of [0, 1] has points in common with at least one of the deleted intervals which means that the union of all these deleted intervals is dense in [0, 1]. Therefore C is nowhere dense. Expand each number x in [0, 1] in a ternary expansion 0.a1 a2 . . . that is, P x = n≥0 3−n an , where an = 0, 1, or 2. Then it is easy to see that C consists of all numbers whose ternary expansion involves only 0 and 2. (The number 1 equals 0.222 . . . .) Thus C is in bijective correspondence with all sequences valued in a two-point set, that is, the cardinality of C is that of the continuum; that is, C is uncountable. We shall construct a C 1 function f : R2 → R which is not C 2 and which contains [0, 2] among its critical values. Since the measure of this set equals 2, this contradicts the conclusion of Sard’s theorem. Note, however, that there is no contradiction with the statement of Sard’s theorem since f is only C 1 . We start the construction by noting that the set C + C = { x + y | x, y ∈ C } equals [0, 2]. The reader can easily be convinced of this fact by expanding every number in [0, 2] in a ternary expansion and solving the resulting undetermined system of infinitely many equations. (The number 2 equals 1.222 . . . .) Assume that we have constructed a C 1 -function g : R → R which contains C among its critical values. The function f (x, y) = g(x)+g(y) is C 1 , and if c1 , c2 ∈ C, then there are critical points x1 , x2 ∈ [0, 1] such that g(xi ) = ci , i = 1, 2; that is, (x1 , x2 ) is a critical point of f and its critical value is c1 + c2 . Since C + C = [0, 2], the set of critical values of f contains [0, 2]. We proceed to the construction of a function g : R → R containing C in its set of critical points. At the kth step in the construction of C, we delete 2k−1 open intervals, each of length 3−k . On these 2k−1 intervals, construct (smooth) congruent bump functions of height 2−k and area = (const.) 2−k 3−k (Figure 3.6.1).

Figure 3.6.1. The construction of congruent bump functions

These define a smooth function hk ; let gk (x) be the integral from −∞ to x of hk , so gk0 = hk and gk is smooth. At each endpoint of the intervals, hk vanishes, that is, the finite set of endpoints occurring in the

3.6 The Sard and Smale Theorems

199

P k-th step of the construction of C is among the critical P points of gk . It is easy to see that h = k≥1 hk is a uniformly convergent Cauchy series and that g = k≥1 gk is pointwise Cauchy; note that gk is monotone and gk (1) − gk (0) = (const. ) 3−k . Therefore, g defines a C 1 function with g 0 = h. The reader can convince themselves that h has arbitrarily steep slopes so that g is not C 2 . The above example was given by Grinberg [1985]. Other examples of this sort are due to Whitney [1935] and Kaufman [1979]. We proceed to the global version of Sard’s theorem on finite-dimensional manifolds. Recall that a subset of a topological space is residual if it is the intersection of countably many open dense sets. The Baire category theorem 1.7.4 asserts that a residual subset of a a locally compact space or of a complete pseudometric space is dense. A topological space is called Lindel¨ of if every open covering has a countable subcovering. In particular, second countable topological spaces are Lindel¨of. (See Lemma 1.1.6.) 3.6.5 Theorem (Sard’s Theorem for Manifolds). Let M and N be finite-dimensional C k manifolds, dim(M ) = m, dim(N ) = n, and f : M → N a C k mapping, k ≥ 1. Assume M is Lindel¨ of and k > max(0, m − n). Then Rf is residual and hence dense in N . Proof. Denote by C the set of critical points of f . We will show that every x ∈ M has a neighborhood Z such that Rf | cl(Z) is open and dense. Then,Tsince M is Lindel¨of we can find a countable cover {Zi } of X with Rf | cl(Zi ) open and dense. Since Rf = i Rf | cl(Zi ), it will follow that Rf is residual. Choose x ∈ M . We want a neighborhood Z of x with Rf | cl(Z) open and dense. By taking local charts we may assume that M is an open subset of Rm and N = Rn . Choose an open neighborhood Z of x such that cl(Z) is compact. Then C = { x ∈ M | rank Df (x) < n } is closed, so cl(Z) ∩ C is compact, and hence f (cl(Z) ∩ C) is compact. But f (cl(Z) ∩ C) is a subset of the set of critical values of f and hence, by Sard’s theorem in Rn , has measure zero. A closed set of measure zero is nowhere dense; hence Rf | cl(Z) = Rn \f (cl(Z) ∩ C) is open and dense.  We leave it to the reader to show that the concept of measure zero makes sense on an n-manifold and to deduce that the set of critical values of f has measure zero in N . Infinite Dimensional Case. To consider the infinite-dimensional version of Sard’s theorem, we first analyze the regular points of a map. 3.6.6 Lemma.

The set SL(E, F) of linear continuous split surjective maps is open in L(E, F).

Proof. Choose A ∈ SL(E, F), write E = F ⊕ K where K is the kernel of A, and define A0 : E → F × K by A0 (e) = (A(e), p(e)) where p : E = F ⊕ K → K is the projection. By the closed graph theorem, p is continuous; hence A0 ∈ GL(E, F × K). Consider the map T : L(E, F × K) → L(E, F) given by T (B) = π ◦ B ∈ L(E, F × K), where π : F × K → F is the projection. Then T is linear, continuous (kπ ◦ Bk ≤ kπk kBk), and surjective; hence, by the open mapping theorem, T is an open mapping. Since GL(E, F × K) is open in L(E, F × K), it follows that T (GL(E, F × K)) is open in L(E, F). But A = T (A0 ) and T (GL(E, F × K)) ⊂ SL(E, F). This shows that SL(E, F) is open.  3.6.7 Proposition. Let f : M → N be a C 1 mapping of manifolds. Then the set of regular points is open in M . Consequently the set of critical points of f is closed in M . Proof. It suffices to prove the proposition locally. Thus, if E, F are the model spaces for M and N , respectively, and x ∈ U ⊂ E is a regular point of f , then Df (x) ∈ SL(E, F). Since Df : U → L(E, F) is continuous, (Df )−1 (SL(E, F)) is open in U by Lemma 3.6.6. 

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3. Manifolds and Vector Bundles

3.6.8 Corollary. Let f : M → N be C 1 and P a submanifold of N . The set { m ∈ M | f is transversal to P at m } is open in M . Proof. Assume f is transversal to P at m ∈ M . Choose a submanifold chart (V, ϕ) at f (m) ∈ P , ϕ : V → F1 × F2 , ϕ(V ∩ P ) = F1 × {0}. Hence if π : F1 × F2 → F2 is the canonical projection, V ∩ P = ϕ−1 (F1 × {0}) = (π ◦ ϕ)−1 {0}. Clearly, π ◦ ϕ : V ∩ P → F2 is a submersion so that by Theorem 3.5.4, ker Tf (m) (π ◦ ϕ) = Tf (m) P . Thus f is transversal to P at the point f (m) iff Tf (p) N = ker Tf (m) (π ◦ ϕ) + Tf (p) P and (Tm f )−1 (Tf (m)P ) = ker Tm (π ◦ ϕ ◦ f ) splits in Tm M . Since ϕ ◦ π is a submersion this is equivalent to π ◦ ϕ ◦ f being submersive at m ∈ M (see Exercise 2.2-5). From Proposition 3.6.7, the set where π ◦ ϕ ◦ f is submersive is open in U , hence in M , where U is a chart domain such that f (U ) ⊂ V .  3.6.9 Example. If M and N are Banach manifolds, the Sard theorem is false without further assumptions. The following counterexample is, so far as we know, due to Bonic, Douady, and Kupka. Let X  xj 2 < ∞ }, E = { x = (x1 , x2 , . . . ) | xi ∈ R, kxk2 = j j≥1

which is P a Hilbert space with respect to the usual algebraic operations on components and the inner product hx, yi = j≥1 xj yj /j 2 . Consider the map f : E → R given by f (x) =

X −2x3j + 3x2j j≥1

2j

,

which is defined since x ∈ E implies |xi | < c for some c > 0 and thus −2x3 + 3x2 2c3 j 3 + 3c2 j 2 c0 j 3 j j ≤ < ; 2j 2j 2j that is, the series f (x) is majorized by the convergent series c0 Df (x) · v =

P

j≥1

X 6(−x2j + xj )vj j≥1

2j

j 3 /2j . We have ;

that is, f is C 1 . In fact f is C ∞ . Moreover, Df (x) = 0 iff all coefficients of vj are zero, that is, iff xj = 0 or xj = 1. Hence the set of critical points is { x ∈ E | xj = 0 or 1 } so that the set of critical values is   ∞ X  xj { f (x) | xj = 0 or xj = 1 } = 0 or x = 1 = [0, 1]. x = j j   sj j=1 But clearly [0, 1] has measure one.



Sard’s theorem holds, however, if enough restrictions are imposed on f . The generalization we consider is due to Smale [1965]. The class of linear maps allowed are Fredholm operators which have splitting properties similar to those in the Fredholm alternative theorem. 3.6.10 Definition. operator if :

Let E and F be Banach spaces and A ∈ L(E, F). Then A is called a Fredholm

3.6 The Sard and Smale Theorems

201

(i) A is double splitting; that is, both the kernel and the image of A are closed and have closed complement; (ii) the kernel of A is finite dimensional ; (iii) the range of A has finite codimension. In this case, if n = dim(ker A) and p = codim(range (A)), index (A) := n − p is the index of A. If M and N are C 1 manifolds and f : M → N is a C 1 map, we say f is a Fredholm map if for every x ∈ M , Tx f is a Fredholm operator. Condition (i) follows from (ii) and (iii); see Exercises 2.2-8 and 2.2-14. A map g between topological spaces is called locally closed if every point in the domain of definition of g has an open neighborhood U such that g| cl (U ) is a closed map (i.e., maps closed sets to closed sets). 3.6.11 Lemma.

A Fredholm map is locally closed.

Proof. By the local representative theorem we may suppose our Fredholm map has the form f (e, x) = (e, η(e, x)), for e ∈ D1 , and x ∈ D2 , where f : D1 × D2 → E × Rp and D1 ⊂ E, D2 ⊂ Rn are open unit balls. Let U1 and U2 be open balls with cl(U1 ) ⊂ D1 and cl(U2 ) ⊂ D2 . Let U = U1 × U2 so that cl(U ) = cl(U1 ) × cl(U2 ). Then f | cl(U ) is closed. To see this, suppose A ⊂ cl(U ) is closed; to show f (A) is closed, choose a sequence {(ei , yi )} such that (ei , yi ) → (e, y) as i → ∞ and (ei , yi ) ∈ f (A), say (ei , yi ) = f (ei , xi ), where (ei , xi ) ∈ A. Since xi ∈ cl(U2 ) and cl(U2 ) is compact, we may assume xi → x ∈ cl(U2 ). Then (ei , xi ) → (e, x). Since A is closed, (e, x) ∈ A, and f (e, x) = (e, y), so (e, y) ∈ f (A). Thus f (A) is closed.  3.6.12 Theorem (The Smale–Sard Theorem). Let M and N be C k manifolds with M Lindel¨ of and assume that f : M → N is a C k Fredholm map, k ≥ 1. Suppose that k > index (Tx f ) for every x ∈ M . Then Rf is a residual subset of N . Proof. It suffices to show that every x0 ∈ M has a neighborhood Z such that R(f |Z) is open and dense in N . Choose z ∈ M . We shall construct a neighborhood Z of z so that R(f |Z) is open and dense. By the local representation theorem we may choose charts (U, α) at z and (V, β) at f (z) such that α(U ) ⊂ E × Rn , β(V ) ⊂ E × Rp and the local representative fαβ = β ◦ f ◦ α−1 of f has the form fαβ (e, x) = (e, η(e, x)) for (e, x) ∈ α(U ). (Here x ∈ Rn , e ∈ E, and η : α(U ) → Rp .) The index of Tz f is n − p and so k > max(0, n − p) by hypothesis. We now show that R(f |U ) is dense in N . Indeed it suffices to show that R(fαβ ) is dense in E × Rp . For e ∈ E, (e, x) ∈ α(U ), define ηe (x) = η(e, x). Then for each e, ηe is a C k map defined on an open set of Rn . By Sard’s theorem, R(ηe ) is dense in Rn for each e ∈ E. But for (e, x) ∈ α(U ) ⊂ E × Rn , we have   I 0 Dfαβ (e, x) = ∗ Dηe (x) so Dfαβ (e, x) is surjective iff Dηe (x) is surjective. Thus for e ∈ E {e} × R(ηe ) = R(fαβ ) ∩ ({e} × Rp ) and so R(fαβ ) intersects every plane {e} × Rp in a dense set and is, therefore, dense in E × Rp , by Lemma 3.6.2. Thus R(f |U ) is dense as claimed. By Lemma 3.6.11 we can choose an open neighborhood Z of z such that cl(Z) ⊂ U and f | cl(Z) is closed. By Proposition 3.6.7 the set C of critical points of f is closed in M . Hence, f (cl(Z) ∩ C) is closed in N and so R(f | cl(Z)) = N \f (cl(Z) ∩ C) is open in N . Since R(f |U ) ⊂ R(f | cl(Z)), this latter set is also dense. We have shown that every point z ∈ N has an open neighborhood Z such that R(f | cl(Z)) is open and dense in N . Repeating the argument of Theorem 3.6.5 shows that Rf is residual (recall that M is Lindel¨of). 

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Sard’s theorem deals with the genericity of the surjectivity of the derivative of a map. We now address the dual question of genericity of the injectivity of the derivative of a map. 3.6.13 Lemma.

The set IL(E, F) of linear continuous split injective maps is open in L(E, F).

Proof. Let A ∈ IL(E, F). Then A(E) is closed and F = A(E) ⊕ G for G a closed subspace of F. The map Γ : E × G → F; defined by Γ(e, g) = A(e) + g is clearly linear, bijective, and continuous, so by Banach’s isomorphism theorem Γ ∈ GL(E × G, F). The map P : L(E × G, F) → L(E, F) given by P (B) = B|E is linear, continuous, and onto, so by the open mapping theorem it is also open. Moreover P (Γ) = A and P (GL(E × G, F)) ⊂ IL(E, F) for if B ∈ GL(E × G, F) then F = B(E) ⊕ B(G) where both B(E) and B(G) are closed in F. Thus A has an open neighborhood P (GL(E × G, F)) contained in IL(E, F).  3.6.14 Proposition.

Let f : M → N be a C 1 -map of manifolds. The set P = { x ∈ M | f is an immersion at x }

is open in M . Proof. It suffices to prove the proposition locally. If E and F are the models of M and N respectively and if f : U → E is immersive at x ∈ U ⊂ E, then Df (x) ∈ IL(E, F). By Lemma 3.6.13, (Df )−1 (IL(E, F)) is open in U since Df : U → L(E, F) is continuous.  The analog of the openness statements in Propositions 3.6.7 and 3.6.14 for subimmersions follows from Definition 3.5.15. Indeed, if f : M → N is a C 1 map which is a subimmersion at x ∈ M , then there is an open neighborhood U of x, a manifold P , a submersion s : U → P , and an immersion j : P → N such that f |U = j ◦ s. But this says that f is subimmersive at every point of U . Thus we have the following. 3.6.15 Proposition.

Let f : M → N be a C 1 -map of manifolds. Then the set P = { x ∈ M | f is a subimmersion at x }

is open in M . If M or N are finite dimensional then P = { x ∈ M | rank Tx f is locally constant } by Proposition 3.5.16. Lower semicontinuity of the rank (i.e., each point x ∈ M admits an open neighborhood of U such that rank Ty f ≥ rank Tx f for all y ∈ U ; see Exercise 2.5-9(i)) implies that P is dense. Indeed, if V is any open subset of M , by lower semicontinuity { x ∈ V | rank Tx f is maximal } is open in V and obviously contained in P . Thus we have proved the following. 3.6.16 Proposition. Let f : M → N be a C 1 -map of manifolds where at least one of M or N are finite dimensional. Then the set P = { x ∈ M | f is a subimmersion at x } is dense in M . 3.6.17 Corollary. Let f : M → N be a C 1 injective map of manifolds and let dim(M ) = m. Then the set P = { x ∈ M | f is immersive at x } is open and dense in M . In particular, if dim(N ) = n, then m ≤ n. Proof. By Propositions 3.6.15 and 3.6.16, it suffices to show that if f : M → N is a C 1 -injective map which is subimmersive at x, then it is immersive at x. Indeed, if f |U = j ◦s where U is an open neighborhood of x on which j is injective, then the submersion s must also be injective. Since submersions are locally onto, this implies that s is a diffeomorphism in a neighborhood of x, that is, f restricted to a sufficiently small neighborhood of x is an immersion.  There is a second proof of this corollary that is independent of Proposition 3.6.16. It relies ultimately on the existence and uniqueness of integral curves of C 1 vector fields. This material will be treated in Chapter 4, but we include this proof here for completeness.

3.6 The Sard and Smale Theorems

Alternative Proof of Corollary 3.6.17. use induction on k to show that

203

(D. Burghelea.) It suffices to work in a local chart V . We shall cl(Ui(1),...,i(k) ) ⊃ V,

where Ui(1),...,i(k) is the set of x ∈ V such that  Tx f

∂ ∂xi(1)



 , . . . , Tx f

∂ ∂xi(k)

 are linearly independent.

The case k = n gives then the statement of the theorem. Note that by the preceding proposition Ui(1),...,i(k) is open in V . The statement is obvious for k = 1 since if it fails Tx f would vanish on an open subset of V and thus f would be constant on V , contradicting the injectivity of f . Assume inductively that the statement for k holds; that is, Ui(1),...,i(k) is open in V and cl(Ui(1),...,i(k) ) ⊃ V . Define 0 Ui(k+1) =



   ∂ x ∈ Ui(1),...,i(k) Tx f = 6 0 ∂xi(k+1)

and notice that it is open in Ui(1),...,i(k) and thus in V . It is also dense in Ui(1),...,i(k) (by the case k = 1) and hence in V by induction. Define the following subset of Ui(1),...,i(k+1) 0 Ui(1),...,i(k+1)

 =

x∈

0 Ui(k+1)

    ∂ ∂ Tx f , . . . , Tx f ∂xi(1) ∂xi(k+1)

are linearly independent } . 0 0 We prove that Ui(1),...,i(k+1) is dense in Ui(k+1) , which then shows that cl(Ui(1),...,i(k+1) ) ⊃ V . If this were 0 not the case, there would exist an open set W ⊂ Ui(k+1) such that

1

a (x)Tx f



∂ ∂xi(1)



k

+ · · · + a (x)Tx f



∂ ∂xi(k)



 + Tx f

∂ ∂xi(k+1)

 =0

for some C 1 -functions a1 , . . . , ak nowhere zero on W . Let c : ]−ε, ε[ → W be an integral curve of the vector field ∂ ∂ ∂ a1 i(1) + · · · + ak i(k) + i(k+1) . ∂x ∂x ∂x Then (f ◦ c)0 (t) = Tc(t) f (c0 (t)) = 0, so f ◦ c is constant on ]−ε, ε[ contradicting injectivity of f .



There is no analogous result for surjective maps known to us; an example of a surjective function R → R which has zero derivative on an open set is given in Figure 3.6.2. However surjectivity of f can be replaced by a topological condition which then yields a result similar to the one in Corollary 3.6.17. 3.6.18 Corollary. Let f : M → N be a C 1 -map of manifolds where dim(N ) = n. If f is an open map, then the set { x ∈ M | f is a submersion at x } is dense in M . In particular, if dim(M ) = m, then m ≥ n. Proof. It suffices to prove that if f is a C 1 -open map which is subimmersive at x, then it is submersive at x. This follows from the relation f |U = j ◦ s and the openness of f and s, for then j is necessarily open and hence a diffeomorphism by Theorem 3.5.7(iii). 

Supplement 3.6A

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3. Manifolds and Vector Bundles

y

0

x

Figure 3.6.2. A surjective function with a zero derivative on an open set

An Application of Sard’s Theorem to Fluid Mechanics The Navier–Stokes equations governing homogeneous incompressible flow in a region Ω in R3 for a velocity field u are ∂u + (u · ∇)u − ν∆u = −∇p + f ∂t div u = 0

in Ω

(3.6.2) (3.6.3)

where u is parallel to ∂Ω (so fluid does not escape) and u=ϕ

on ∂Ω.

(3.6.4)

Here, f is a given external forcing function assumed divergence free, p is the pressure (unknown), ϕ is a given boundary condition and ν is the viscosity. Stationary solutions are defined by setting ∂u/∂t = 0. Given f , ϕ and ν the set of possible stationary solutions u is denoted S(f , ϕ, ν). A theorem of Foias and Temam [1977] states (amongst other things) that there is an open dense set O in the Banach space of all (f , ϕ)’s such that S(f , ϕ, ν) is finite for each (f , ϕ) ∈ O. We refer the reader to the cited paper for the precise (Sobolev) spaces needed for f , ϕ, u, and rather give the essential idea behind the proof. Let E be the space of possible u’s (of class H 2 , div u = 0 and u parallel to ∂Ω), F the product of the space H of (L2 ) divergence free vector fields with the space of vector fields on ∂Ω (of class H exc:3.2−27 ). We can rewrite the equation (u · ∇)u − ν∆u = −∇p + f

(3.6.5)

νAu + B(u) = f

(3.6.6)

as where Au = −PH ∆u, PH being the orthogonal projection to H (this is a special instance of the Hodge decomposition; see §7.5 for details) and B(u) = PH ((u · ∇)u). The orthogonal projection operator really encodes the pressure term. Effectively, p is solved for by taking the divergence of (3.6.4) to give ∆p in terms of u and the normal component of (3.6.4) gives the normal derivative of p. The resulting Neuman problem is solved, thereby eliminating p from the problem. Define the map Φν : E → F by Φν (u) = (νAu + B(u), u|∂Ω). One shows that Φν is a C ∞ map by using the fact that A is a bounded linear operator and B is obtained from a continuous bilinear operator; theorems about Sobolev spaces are also required here. Moreover, elliptic

3.6 The Sard and Smale Theorems

205

theory shows that the derivative of Φν is a Fredholm operator, so Φν is a Fredholm map. In fact, from self adjointness of A and DB(u), one sees that Φν has index zero. The Sard–Smale theorem shows that the set of regular values of Φν forms a residual set Oν . It is easy to see that Oν = O is independent of ν. Now since Φν has index zero, at a regular point, DΦν is an isomorphism, so Φν is a local diffeomorphism. Thus we conclude that S(f , ϕ, ν) is discrete and that O is open (Foias and Temam [1977] give a direct proof of openness of O rather than using the implicit function theorem). One knows, also from elliptic theory that S(f , ϕ, ν) is compact, so being discrete, it is finite. One can also prove a similar generic finiteness result for an open dense set of boundaries ∂Ω using a transversality analogue of the Smale–Sard theorem (see Supplement 3.6B), as was pointed out by A. J. Tromba. We leave the precise formulation as a project for the reader.

Supplement 3.6B The Parametric Transversality Theorem 3.6.19 Theorem (Density of Transversal Intersection). Let P, M, N be C k manifolds, S ⊂ N a submanifold (not necessarily closed ) and F : P × M → N a C k map, k ≥ 1. Assume (i) M is finite dimensional (dim M = m) and that S has finite codimension q in N . (ii) P × M is Lindel¨ of. (iii) k > max(0, n − q). (iv) F t S. Then t (F, S) := { p ∈ P | Fp : M → N is transversal to S at all points of S } is residual in P . The idea is this. Since F t S, F −1 (S) ⊂ P × M is a submanifold. The projection π : F −1 (S) → P has the property that a value of π is a regular value iff Fp is transverse to S. We then apply Sard’s theorem to π. A main application is this: consider a family of perturbations f : ]−1, 1[ × M → N of a given map f0 : M → N , where f (0, x) = f0 (x). Suppose f t S. Then there exist t’s arbitrarily close to zero such that ft t S; that is, slight perturbations of f0 are transversal to S. For the proof we need two lemmas. 3.6.20 Lemma. Let E and F be Banach spaces, dim F = n, pr1 : E × F → E the projection onto the first factor, and G ⊂ E × F a closed subspace of codimension q. Denote by p the restriction of pr1 to G. Then p is a Fredholm operator of index n − q. Proof.

Let H = G + ({0} × F)

and K = G ∩ ({0} × F).

Since F is finite dimensional and G is closed, it follows that H is closed in E × F (see Exercise 2.213(ii)). Moreover, H has finite codimension since it contains the finite-codimensional subspace G. Therefore H is split (see Exercise 2.2-14) and thus there exists a finite-dimensional subspace S ⊂ E × {0} such that E×F = H⊕S. Since K ⊂ F, choose closed subspaces G0 ⊂ G and F0 ⊂ {0}×F such that G = G0 ⊕K and {0}×F = K⊕F0 . Thus H = G0 ⊕K⊕F0 and E×F = G0 ⊕K⊕F0 ⊕S. Note that pr1 |G0 ⊕S : G0 ⊕S → E is an isomorphism, K = ker p, and pr1 (S) is a finite-dimensional complement to p(G) in F. Thus p is a Fredholm operator and its index equals dim(K) − dim(S) = dim(K ⊕ F0 ) − dim(S ⊕ F0 ). Since K ⊕ F0 = {0} × F and F0 ⊕ S is a complement to G in E × F (having therefore dimension q by hypothesis), the index of p equals n − q. 

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3.6.21 Lemma. In the hypothesis of Theorem 3.6.19, let V = F −1 (S). Let π 0 : P × M → P be the projection onto the first factor and let π = π 0 |V . Then π is a C k Fredholm map of constant index n − q. Proof. By Theorem 3.6.19(iv), V is a C k submanifold of P × M so that π is a C k map. The map T(p,m) π : T(p,m) V → Tp P is Fredholm of index n − q by Lemma 3.6.20: E is the model of P , F the model of M , and G the model of V .  Proof of Theorem 3.6.19. We shall prove below that p is a regular value of π if and only if Fp t S. If this is shown, since π : V → P is a C k Fredholm map of index n − q by Lemma 3.6.21, the codimension of V in E × F equals the codimension of S in N which is q, k > max(0, n − q), and V is Lindel¨of as a closed subspace of the Lindel¨ of space P × M , the Smale–Sard theorem 3.6.12 implies that t (F, S) is residual in P. By definition, (iv) is equivalent to the following statement: (a) For every (p, m) ∈ P × M satisfying F (p, m) ∈ S, T(m,p) F (Tp P × Tm M ) + TF (p,m) S = TF (p,m) N and (T(m,p) F )−1 (TF (p,m) S) splits in Tp P × Tm M . Since M is finite dimensional, the map m ∈ M 7→ F (p, m) ∈ N for fixed p ∈ P is transversal to S if and only if (b) for every m ∈ M satisfying F (p, m) ∈ S, Tm Fp (Tm M ) + TF (p,m) S = TF (p,m) S. Since π is a Fredholm map, the kernel of T π at any point splits being finite dimensional (see Exercise 2.2-14). Therefore p is a regular value of π if and only if (c) for every m ∈ M satisfying F (p, m) ∈ S and every v ∈ Tp P , there exists u ∈ Tm M such that T(m,p) F (v, u) ∈ TF (p,m) S. We prove the equivalence of (b) and (c). First assume (c), take m ∈ M , p ∈ P such that F (p, m) ∈ S and let w ∈ TF (p,m) S. By (a) there exists v ∈ Tp P , u1 ∈ Tm M , z ∈ TF (p,m) S such that T(m,p) F (v, u1 ) + z = w. By (c) there exists u2 ∈ Tm M such that T(m,p) F (v, u2 ) ∈ TF (p,m) S. Therefore, w = T(m,p) F (v, u1 ) − T(m,p) F (v, u2 ) + T(m,p) F (v, u2 ) + z = T(m,p) F (0, u1 − u2 ) + T(m,p) F (v, u2 ) + z = T(m,p) F (0, u) + z 0 ∈ Tm Fp (Tm M ) + TF (p,m) S, where u = u1 − u2 and z 0 = T(m,p) F (v, u2 ) + z ∈ TF (p,m) S. Thus (b) holds. Conversely, let (b) hold, take p ∈ P , m ∈ M such that F (p, m) ∈ S and let v ∈ Tp P . Pick u1 ∈ Tm M , z1 ∈ TF (p,m) S and define w = T(m,p) F (v, u1 ) + z1 . By (b), there exist u2 ∈ Tm M and z2 ∈ TF (p,m) S such that w = Tm Fp (u2 ) + z2 . Subtract these two relations to get 0 = T(m,p) F (v, u1 ) − Tm Fp (u2 ) + z1 − z2 = T(m,p) F (v, u1 − u2 ) + z1 − z2 , that is, T(m,p) F (v, u1 − u2 ) = z2 − z1 ∈ TF (m,p) S and therefore (c) holds.



There are many other very useful theorems about genericity of transversal intersection. We refer the reader to Golubitsky and Guillemin [1974] and Hirsch [1976] for the finite dimensional results and to Abraham and Robbin [1967] for the infinite dimensional case and the situation when P is a manifold of maps.

3.6 The Sard and Smale Theorems

207

Exercises  3.6-1. Construct a C ∞ function f : R → R whose set of critical points equals [0, 1]. This shows that the set of regular points is not dense in general.  3.6-2. Construct a C ∞ function f : R → R which has each rational number as a critical value. Hint: Since Q is countable, write it as a sequence { qn | n = 0, 1, . . . }. Construct on the closed interval [n, n + 1] a C ∞ function which is zero near n and n + 1 and equal to qn on an open interval. Define f to equal fn on [n, n + 1].  3.6-3.

Show that if m < n there is no C 1 map of an open set of Rm onto an open set of Rn .

 3.6-4. A manifold M is called C k -simply connected , if it is connected and if every C k map f : S 1 → M is C k -homotopic to a constant, that is, there exist a C k -map H : ]−ε, 1 + ε[ × S 1 → M such that for all s ∈ S 1 , H(0, s) = f (s) and H(1, s) = m0 , where m0 ∈ M . (i) Show that the sphere S n , n ≥ 2, is C k -simply connected for any k ≥ 1. Hint: By Sard, there exists a point x ∈ S n \f (S 1 ). Then use the stereographic projection defined by x. (ii) Show that S n , n ≥ 2, is C 0 -simply connected. Hint: Approximate the continuous map g : S 1 → S n by a C 1 -map f : S 1 → S n . Show that one can choose f to be homotopic to g. (iii) Show that S 1 is not simply connected.  3.6-5. Let M and N be submanifolds of Rn . Show that the set { x ∈ Rn | M intersects N +x transversally } is dense in Rn . Find an example when it is not open.  3.6-6. set

Let f : Rn → R be C 2 and consider for each a ∈ Rn the map fa (x) = f (x) + a · x. Prove that the { a ∈ Rn | the matrix [∂ 2 fa (x0 )/∂xi ∂xj ] is nonsingular for every critical point x0 of fa }

is a dense set in Rn which is a countable intersection of open sets. Hint: Use Supplement 3.6B; when is the map (a, x) 7→ ∇f (x) + a transversal to {0}?  3.6-7. Let M be a C 2 manifold and f : M → R a C 2 map. A critical point m0 of f is called nondegenerate, if in a local chart (U, ϕ) at m0 , ϕ(m0 ) = 0, ϕ : U → E, the bilinear continuous map D2 (f ◦ ϕ)−1 (0) : E × E → R is strongly non-degenerate, that is, it induces an isomorphism of E with E∗ . (i) Show that the notion of non-degeneracy is chart independent. Functions all of whose critical points are nondegenerate are called Morse functions. (ii) Assume M is a C 2 submanifold of Rn and f : M → R is a C 2 function. For a ∈ Rn define fa : M → R by fa (x) = f (x) + a · x. Show that the set { a ∈ Rn | fa is a Morse function } is a dense subset of Rn which is a countable intersection of open sets. Show that if M is compact, this set is open in Rn . Hint: Show first that if dim M = m and (x1 , . . . , xn ) are the coordinates of a point x ∈ M in Rn , there is a neighborhood of x in Rn such that m of these coordinates define a chart on M . Cover M with countably many such neighborhoods. In such a neighborhood U , consider the function g : U ⊂ Rn → R defined by g(x) = f (x) + am+1 xm+1 + · · · + an xn . Apply Exercise 3.6-6 to the map fa0 (x) = g(x) + a1 x1 + · · · + am xm , a0 = (a1 , . . . , am ) and look at the set S = { a ∈ Rn | fa is not Morse on U }. Consider S ∩ (Rm × {am+1 , . . . , an }) and apply Lemma 3.6.2.

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(iii) Assume M is a C 2 -submanifold of Rn . Show that there is a linear map α : Rn → R such that α|M is a Morse function. (iv) Show that the “height functions” on S n and Tn are Morse functions.  3.6-8. Let E and F be Banach spaces. A linear map T : E → F is called compact if it maps bounded sets into relatively compact sets. (i) Show that a compact map is continuous. (ii) Show that the set K(E, F) of compact linear operators from E to F is a closed subspace of L(E, F). (iii) If G is another Banach space, show that L(F, G) ◦ K(E, F) ⊂ K(E, G), and that K(E, F) ◦ L(G, E) ⊂ K(G, F). (iv) Show that if T ∈ K(E, F), then T ∗ ∈ K(F∗ , E∗ ).  3.6-9 (F. Riesz). Show that if K ∈ K(E, F) where E and F are Banach spaces and a is a scalar (real or complex), then T = Identity +aK is a Fredholm operator. Hint: It suffices to prove the result for a = −1. Show ker T is a locally compact space by proving that K(D) = D, where D is the open unit ball in ker T . To prove that T (E) is closed and finite dimensional show that dim(E/T (E)) = dim(E/T (E))∗ = dim(ker T ∗ ) = dim(ker(Identity −K ∗ )) < ∞ and use Exercise 2.2-8.  3.6-10. Show that there exist Fredholm operators of any index. Hint: Consider the shifts (x1 , x2 , . . . ) 7→ (0, . . . , 0, x1 , x2 )

and

(x1 , x2 . . . ) 7→ (xn , xn+1 , . . . )

in `2 (R).  3.6-11. Show that if T ∈ L(E, F) is a Fredholm operator, then T ∗ ∈ L(F∗ , E∗ ) is a Fredholm operator and index (T ∗ ) = − index(T ).  3.6-12. (i) Let E, F, G be Banach spaces and T ∈ L(E, F). Assume that there are S, S 0 ∈ L(F, E) such that S ◦ T − Identity ∈ K(E, E) and T ◦ S 0 − Identity ∈ K(F, F). Show that T is Fredholm. Hint: Use Exercise 3.6-9. (ii) Use (i) to prove that T ∈ L(E, F) is Fredholm if and only if there exists an operator S ∈ L(F, E) such that (S ◦ T − Identity) and (T ◦ S− Identity) have finite dimensional range. Hint: If T is Fredholm, write E = ker T ⊕ F0 , F = T (E) ⊕ F0 and show that T0 = T |E0 : E0 → T (E) is a Banach space isomorphism. Define S ∈ L(F, E) by S|T (E) = T0−1 , S|F0 = 0. (iii) Show that if T ∈ L(E, F), K ∈ K(E, F) then T + K is Fredholm. (iv) Show that if T ∈ L(E, F), S ∈ L(F, G) are Fredholm, then so is S ◦ T and that index(S ◦ T ) = index(S) + index(T ).  3.6-13.

Let E, F be Banach spaces.

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209

(i) Show that the set Fredq (E, F) = { T ∈ L(E, F) | T is Fredholm, index(T ) = q } is open in L(E, F). Hint: Write E = ker T ⊕ E0 , F = T (E) ⊕ F0 and define T˜ : E ⊕ F0 → F ⊕ ker T by T˜(z ⊕ x, y) = (T (x) ⊕ y, z), for x ∈ E0 , z ∈ ker T , y ∈ F0 . Show that T˜ ∈ GL(E ⊕ F0 , F ⊕ ker T ). Define ρ : L(E ⊕ F0 , F ⊕ ker T ) → L(E, F) by ρ(S) = π ◦ S ◦ i, where π : F ⊕ ker T → F is the projection and i : e ∈ E 7→ (e, 0) ∈ E ⊕ F0 is the inclusion. Show that ρ is a continuous linear surjective map and hence open. Prove ρ(GL(E ⊕ F0 , F ⊕ ker T ) ⊂ Fredq (E, F), ρ(T˜) = T. (ii) Conclude from (i) that the index map from Fred(E, F) to Z is constant on each connected component of Fred(E, F) = { T ∈ L(E, F) | T is Fredholm }. Show that if E = F = `2 (R) and T (t)(x1 , x2 , . . . ) = (0, tx2 , x3 , . . . ) then index(T (t)) equals 1, but dim(ker(T (t))) and dim(`2 (R)/T (t)(`2 (R))) jump at t = 0. (iii) (Homotopy invariance of the index.) Show that if ϕ : [0, 1] → Fred(E, F) is continuous, then index(ϕ(0)) = index(ϕ(1)). Hint: Let a = sup{ t ∈ [0, 1] | s < t implies index(f (s)) = index(f (0)) }. By (i) we can find ε > 0 such that |b − a| < ε implies index(f (b)) = index(f (a)). Let b = a − ε/2 and thus index(f (0)) = index(f (b)) = index(f (a)). Show by contradiction that a = 1. (iv) If T ∈ Fred(E, F), K ∈ K(E, F), show that index(T + K) = index(T ). Hint: T + K(E, F) is connected; use (ii). (v) (The Fredholm alternative.) Let K ∈ K(E, F) and a 6= 0. Show that the equation K(e) = ae has only the trivial solution iff for any v ∈ E, there exists u ∈ E such that K(u) = au + v. Hint: K − a(Identity) is injective iff (1/a)K − (Identity) is injective. By (iv) this happens iff (1/a)K − (Identity) is onto.  3.6-14. Using Exercise 3.5-2, show that the map π : SL(Rm , Rk ) × IL(Rk , Rn ) → St(m, n; k), where k ≤ min(m, n), defined by π(A, B) = B ◦ A, is a smooth locally trivial fiber bundle with typical fiber GL(Rk ).  3.6-15. (i) Let M and N be smooth finite-dimensional manifolds and let f : M → N be a C 1 bijective immersion. Show that f is a C 1 diffeomorphism. Hint: If dim M < dim N , then f (M ) has measure zero in N , so f could not be bijective. (ii) Formulate an infinite-dimensional version of (i).

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4 Vector Fields and Dynamical Systems

This version: January 21, 2007

This chapter studies vector fields and the dynamical systems they determine. The ensuing chapters will study the related topics of Lie groups, tensors and differential forms. A basic operation introduced in this chapter is the Lie derivative of a function or a vector field. It is introduced in two different ways, algebraically as a type of directional derivative and dynamically as a rate of change along a flow. The Lie derivative formula asserts the equivalence of these two definitions. The Lie derivative is a basic operation used extensively in differential geometry, general relativity, Hamiltonian mechanics, and continuum mechanics.

4.1

Vector Fields and Flows

This section introduces vector fields and the flows they determine. This topic puts together and globalizes two basic ideas we learn in undergraduate calculus: the study of vector fields on the one hand and differential equations on the other. 4.1.1 Definition. Let M be a manifold. A vector field on M is a section of the tangent bundle T M of M . The set of all C r vector fields on M is denoted by Xr (M ) and the C ∞ vector fields by X∞ (M ) or X(M ). Thus, a vector field X on a manifold M is a mapping X : M → T M such that X(m) ∈ Tm M for all m ∈ M . In other words, a vector field assigns to each point of M a vector based (i.e., bound) at that point. 4.1.2 Example. Consider the force field determined by Newton’s law of gravitation. Here the manifold is R3 minus the origin and the vector field is F(x, y, z) = −

mM G r, r3

where m is the mass of a test body, M is the mass of the central body, G is the constant of gravitation, r is the vector from the origin to (x, y, z), and r = (x2 + y 2 + z 2 )1/2 ; see Figure 4.1.1.  The Concept of a Flow. The study of dynamical systems, also called flows, may be motivated as follows. Consider a physical system that is capable of assuming various “states” described by points in a set S. For

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Figure 4.1.1. The gravitational vector field.

example, S might be R3 × R3 and a state might be the position and momentum (q, p) of a particle. As time passes, the state evolves. If the state is s0 ∈ S at time λ and this changes to s at a later time t, we set Ft,λ (s0 ) = s and call Ft,λ the evolution operator ; it maps a state at time λ to what the state would be at time t; that is, after a time interval t − λ has elapsed. “Determinism” is expressed by the law Fτ,t ◦ Ft,λ = Fτ,λ ,

Ft,t = identity,

sometimes called the Chapman–Kolmogorov law . The evolution laws are called time independent when Ft,λ depends only on t − λ; that is, Ft,λ = Fs,µ

if

t − λ = s − µ.

Setting Ft = Ft,0 , the preceding law becomes the group property : Ft ◦ Fτ = Ft+τ ,

F0 = identity .

We call such an Ft a flow and Ft,λ a time-dependent flow , or as before, an evolution operator. If the system is nonreversible, that is, defined only for t ≥ λ, we speak of a semi-flow . It is usually not Ft,λ that is given, but rather the laws of motion. In other words, differential equations are given that we must solve to find the flow. These equations of motion have the form ds = X(s), dt

s(0) = s0

where X is a (possibly time-dependent) vector field on S. 4.1.3 Example. The motion of a particle of mass m under the influence of the gravitational force field in Example 4.1.2 is determined by Newton’s second law: m

d2 r = F; dt2

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that is, by the ordinary differential equations mM Gx d2 x =− ; dt2 r3 d2 y mM Gy m 2 =− ; dt r3 mM Gz d2 z . m 2 =− dt r3

m

Letting q = (x, y, z) denote the position and p = m(dr/dt) the momentum, these equations become dq p = ; dt m

dp = F(q). dt

The phase space here is the manifold (R3 \{0}) × R3 , that is, the cotangent manifold of R3 \{0}. The righthand side of the preceding equations define a vector field on this six-dimensional manifold by   p , F(q) . X(q, p) = (q, p), m In courses on mechanics or differential equations, it is shown how to integrate these equations explicitly, producing trajectories, which are planar conic sections. These trajectories comprise the flow of the vector field.  Vector Fields and Integral Curves on Manifolds. Let us now turn to the elaboration of these ideas when a vector field X is given on a manifold M . If M = U is an open subset of a Banach space E, then a vector field on U is a map X : U → U × E of the form X(x) = (x, V (x)). We call V the principal part of X. However, having a separate notation for the principal part turns out to be an unnecessary burden. By abuse of notation, in linear spaces we shall write a vector field simply as a map X : U → E and shall mean the vector field x 7→ (x, X(x)). When it is necessary to be careful with the distinction, we shall be. If M is a manifold and ϕ : U ⊂ M → V ⊂ E is a local coordinate chart for M , then a vector field X on M induces a vector field Xlr on E called the local representative of X by the formula Xlr (x) = T ϕ·X(ϕ−1 (x)). The principal part of Xlr is denoted Xloc . Thus, Xlr (x) = (x, Xloc (x)). and so if the domain of definition of the local representative is U ⊂ E, we have Xloc : U ⊂ E → E. In the case E = Rn , we can identify Xloc with an n-component vector function which we denote (X 1 (x), . . . , X n (x)). Thus, we sometimes just say “the vector field X whose local representative is (X 1 , . . . , X n ).” Recall that a curve c at a point m of a manifold M is a C 1 -map from an open interval I of R into M such that 0 ∈ I and c(0) = m. For such a curve we may assign a tangent vector at each point c(t), t ∈ I, by c0 (t) = Tt c(1). 4.1.4 Definition. Let M be a manifold and X ∈ X(M ). An integral curve of X at m ∈ M is a curve c at m such that c0 (t) = X(c(t)) for each t ∈ I. In case M = U ⊂ E, we identify a vector field with its principal part, so that X : U → E. In this case, a curve c(t) is an integral curve of X when c0 (t) = X(c(t)), where c0 = dc/dt. If X is a vector field on a manifold M and Xloc denotes its local representative in a chart ϕ, the chain rule shows that a curve c on M is an integral curve of X iff dcloc (t) = Xloc (c(t)), dt

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where cloc = ϕ ◦ c is the local representative of the curve c. If M is an n-manifold and the local representatives of X and c are (X 1 , . . . , X n ) and (c1 , . . . , cn ) respectively, then c is an integral curve of X iff the following system of ordinary differential equations is satisfied dc1 (t) = X 1 (c1 (t), . . . , cn (t)); dt .. .. . . dcn (t) = X n (c1 (t), . . . , cn (t)). dt The reader should chase through the definitions to verify this assertion. The preceding equations are autonomous, corresponding to the fact that X is time independent. If X were time dependent, time t would appear explicitly on the right-hand side. As we saw in Example 4.1.3, the preceding system of equations includes equations of higher order (by their usual reduction to first-order systems) and the Hamilton equations of motion as special cases. Local Existence and Uniqueness. We now present the local existence and uniqueness theorems for integral curves of vector fields. Using a chart and the above remarks, we see that the crux of the matter concerns these results in a Banach space, so we begin with this setting. 4.1.5 Theorem (Local Existence, Uniqueness, and Smoothness). open, and suppose X : U ⊂ E → E is of class C k , k ≥ 1. Then

Let E be a Banach space, U ⊂ E be

(i) For each x0 ∈ U , there is a curve c : I → U at x0 such that c0 (t) = X(c(t)) for all t ∈ I. (ii) Any two such curves are equal on the intersection of their domains. (iii) There is a neighborhood U0 of the point x0 ∈ U , a real number a > 0, and a C k mapping F : U0 ×I → E, where I is the open interval ]−a, a[, such that the curve cu : I → E, defined by cu (t) = F (u, t) is a curve at u ∈ E satisfying the differential equations c0u (t) = X(cu (t)) for all t ∈ I. Actually the crucial assumption on the vector field is that it be Lipschitz, as in the following preparatory lemma. 4.1.6 Lemma. Let E be a Banach space, U ⊂ E an open set, and X : U ⊂ E → E a Lipschitz map; that is, there is a constant K > 0 such that kX(x) − X(y)k ≤ Kkx − yk for all x, y ∈ U . Let x0 ∈ U and suppose the closed ball of radius b, Bb (x0 ) = { x ∈ E | kx − x0 k ≤ b } lies in U , and kX(x)k ≤ M for all x ∈ Bb (x0 ). Let t0 ∈ R and let α = b/M . Then there is a unique C 1 curve x(t), t ∈ [t0 − α, t0 + α] such that x(t) ∈ Bb (x0 ) and x0 (t) = X(x(t)), Proof.

x(t0 ) = x0 .

The conditions x0 (t) = X(x(t)), x(t0 ) = x0 are equivalent to the integral equation Z t x(t) = x0 + X(x(s)) ds. t0

Put x0 (t) = x0 and define inductively Z

t

xn+1 (t) = x0 +

X(xn (s)) ds t0

(4.1.1)

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Assuming that xn (t) ∈ Bb (x0 ), the following estimate shows that xn+1 (t) ∈ Bb (x0 ). We shall also assume that t ≥ t0 for simplicity. We have Z t kxn+1 (t) − x0 k ≤ kX(xn (s))k ds ≤ (t − t0 )M ≤ αM = b. t0

Thus, by induction, xn (t) ∈ Bb (x0 ) for all n = 1, 2, . . . and for all t ∈ [t0 − α, t0 + α]. We also claim by induction that M Kn |t − t0 |n+1 . kxn+1 (t) − xn (t)k ≤ (n + 1)! To see this, assume that M K n−1 |t − t0 |n . n! Then we estimate as follows, again assuming that t ≥ t0 for simplicity:

Z t



kxn+1 (t) − xn (t)k = (X(xn (s)) − X(xn−1 (s))) ds

t0 Z t ≤ K kxn (s) − xn−1 (s)k ds kxn (t) − xn−1 (t)k ≤

t0

Z

t

M K n−1 (s − t0 )n ds n!

≤K t0

M Kn = (t − t0 )n+1 . (n + 1)! Thus, since

M Kn M K n n+1 (t − t0 )n+1 ≤ α (n + 1)! (n + 1)!

and the series with these quantities as terms is convergent, we see, writing kxn+p − xn k as a telescoping sum, that the functions xn (t) form a uniformly Cauchy sequence and hence converge uniformly to a continuous function x(t). This curve x(t) satisfies the integral equation (4.1.1). Since x(t) is continuous, the integral equation in fact shows that it is C 1 from the fundamental theorem of Calculus. This proves existence. For uniqueness, let y(t) be another solution. By a similar induction argument as above, we find that kxn (t) − y(t)k ≤ M K n |t − t0 |n+1 /(n + 1)!; thus, letting n → ∞ gives x(t) = y(t).  The same argument holds if X depends explicitly on t or on a parameter ρ, is jointly continuous in (t, ρ, x), and is Lipschitz in x uniformly in t and ρ. Since xn (t) is continuous in (x0 , t0 , ρ) so is x(t), being a uniform limit of continuous functions; thus the integral curve is jointly continuous in (x0 , t0 , ρ). 4.1.7 Proposition (Gronwall’s Inequality). Let f, g : [a, b[ → R be continuous and nonnegative. Suppose that for all t satisfying a ≤ t < b, Z t f (t) ≤ A + f (s) g(s) ds, for a constant A ≥ 0. a

Then Z f (t) ≤ A exp

t

 g(s) ds

for all t ∈ [a, b[.

a

Proof.

First suppose A > 0. Let Z h(t) = A +

t

f (s) g(s) ds; a

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thus h(t) > 0. Then h0 (t) = f (t)g(t) ≤ h(t)g(t). Thus h0 (t)/h(t) ≤ g(t). Integration gives Z h(t) ≤ A exp

t

 g(s) ds .

a

This gives the result for A > 0. If A = 0, then we get the result by replacing A by ε > 0 for every ε > 0; thus h and hence f is zero.  4.1.8 Lemma. Let X be as in Lemma 4.1.6. Let Ft (x0 ) denote the solution (= integral curve) of x0 (t) = X(x(t)), x(0) = x0 . Then there is a neighborhood V of x0 and a number ε > 0 such that for every x1 ∈ V there is a unique integral curve x1 (t) = Ft (x1 ) satisfying x01 (t) = X(x1 (t)) for all t ∈ [−ε, ε], and x1 (0) = x1 . Moreover, for any two points x1 , x2 ∈ V , we have kFt (x1 ) − Ft (x2 )k ≤ eK|t| kx1 − x2 k.

(4.1.2)

Proof. Choose V = Bb/2 (x0 ) and ε = b/2M . Let x1 ∈ V . Then Bb/2 (x1 ) ⊂ Bb (x0 ) and hence kX(z)k ≤ M for all z ∈ Bb/2 (y). By Lemma 4.1.6 with x0 replaced by x1 , b by b/2, and t0 by 0, there exists an integral curve x1 (t) of x01 (t) = X(x1 (t)) for t ∈ [−ε, ε] and satisfying x1 (0) = x1 . This proves the first part. For the second part, let f (t) = kFt (x1 ) − Ft (x2 )k. Clearly

Z t

Z t

f (t) = [X(F (x )) − X(F (x ))] ds + x − x ≤ kx − x k + K f (s) ds, s 1 s 2 1 2 1 2

0

0

so the result follows from Gronwall’s inequality.



This result shows that Ft (x) depends in a continuous, indeed Lipschitz, manner on the initial condition x and is jointly continuous in (t, x). Again, the same result holds if X depends explicitly on t and on a parameter ρ, is jointly continuous in (t, ρ, x), and is Lipschitz in x uniformly in t and ρ; (Ft,λ )ρ (x) is the unique integral curve x(t) satisfying x0 (t) = X(x(t), t, ρ) and x(λ) = x. By the remark following Lemma 4.1.6, (Ft,λ )ρ (x) is jointly continuous in the variables (λ, t, ρ, x), and is Lipschitz in x, uniformly in (λ, t, ρ). The next result shows that Ft is C k if X is, which will complete the proof of Theorem 4.1.5. Note that since a C 1 -function is locally Lipschitz, we can apply the above local existence and uniqueness results. 4.1.9 Lemma. Let X in Lemma 4.1.6 be of class C k , l ≤ k ≤ ∞, and let Ft (x) be defined as before. Then locally in (t, x), Ft (x) is of class C k in x and is C k+1 in the t-variable.

Proof. We define ψ(t, x) ∈ L(E, E), the set of continuous linear maps of E to E, to be the solution of the “linearized” or “first variation” equations: d ψ(t, x) = DX(Ft (x)) ◦ ψ(t, x), dt

with ψ(0, x) = identity,

where DX(y) : E → E is the derivative of X taken at the point y. Since the vector field ψ 7→ DX(Ft (x)) ◦ ψ on L(E, E) (depending explicitly on t and on the parameter x) is Lipschitz in ψ, uniformly in (t, x) in a neighborhood of every (t0 , x0 ), by the remark following Lemma 4.1.8 it follows that ψ(t, x) is continuous in (t, x) (using the norm topology on L(E, E)).

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We claim that DFt (x) = ψ(t, x). To show this, fix t, set θ(s, h) = Fs (x + h) − Fs (x), and write t

Z

{X(Fs (x + h)) − X(Fs (x))} ds Z t − [DX(Fs (x)) ◦ ψ(s, x)] · h ds

θ(t, h) − ψ(t, x) · h =

0

0 t

Z =

DX(Fs (x)) · [θ(s, h) − ψ(s, x) · h] ds Z t + {X(Fs (x + h)) − X(Fs (x))

0

0

− DX(Fs (x)) · [Fs (x + h) − Fs (x)]} ds. Since X is of class C 1 , given ε > 0, there is a δ > 0 such that khk < δ implies the second term is dominated in norm by Z t εkFs (x + h) − Fs (x)k ds, 0

which is, in turn, smaller than Aεkhk for a positive constant A by Lemma 4.1.8. By Gronwall’s inequality we obtain kθ(t, h) − ψ(t, x) · hk ≤ (constant) εkhk. It follows that DFt (x) · h = ψ(t, x) · h. Thus both partial derivatives of Ft (x) exist and are continuous; therefore Ft (x) is of class C 1 . We prove Ft (x) is C k by induction on k. Begin with the equation defining Ft : d Ft (x) = X(Ft (x)) dt so

d d Ft (x) = DX(Ft (x)) · X(Ft (x)) dt dt

and

d DFt (x) = DX(Ft (x)) · DFt (x). dt

Since the right-hand sides are C k−1 , so are the solutions by induction. Thus F itself is C k .



Again there is an analogous result for the evolution operator (Ft,λ )ρ (x) for a time-dependent vector field X(x, t, ρ), which depends on extra parameters ρ in a Banach space P. If X is C k , then (Ft,λ )ρ (x) is C k in all variables and is C k+1 in t and λ. The variable ρ can be easily dealt with by suspending X to a new vector field obtained by appending the trivial differential equation ρ0 = 0; this defines a vector field on E × P and Theorem 4.1.5 may be applied to it. The flow on E × P is just Ft (x, ρ) = (Ftρ (x), ρ). For another more “modern” proof of Theorem 4.1.5 see Supplement 4.1C. That alternative proof has a technical advantage: it works easily for other types of differentiability assumptions on X or on Ft , such as H¨older or Sobolev differentiability; this result is due to Ebin and Marsden [1970]. Global Uniqueness. globally.

As we saw above, integral curves are locally unique. Now we prove uniqueness

4.1.10 Proposition (Global Uniqueness). Let M be a Hausdorff manifold and let X be a C k vector field on M , k ≥ 1. Suppose that c1 and c2 are two integral curves of X at the same point m ∈ M . Then c1 = c2 on the intersection of their domains. Proof. This does not follow at once from Theorem 4.1.5 for c1 and c2 may lie in different charts. (Indeed, if the manifold is not Hausdorff, Exercise 4.1-13 shows that this proposition is false.) Suppose c1 : I1 → M and c2 : I2 → M . Let I = I1 ∩ I2 , and let K = { t ∈ I | c1 (t) = c2 (t) }; K is closed since M is Hausdorff. We

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will now show that K is open. From Theorem 4.1.5, K contains some neighborhood of 0. For t ∈ K consider ct1 and ct2 , where ct (s) = c(t + s). Then ct1 and ct2 are integral curves at c1 (t) = c2 (t). Again, by Theorem 4.1.5 they agree on some neighborhood of 0. Thus some neighborhood of t lies in K, and so K is open. Since I is connected, K = I.  Flow Boxes. The mapping F gives a locally unique integral curve cu for each u ∈ U0 , and for each t ∈ I, Ft = F |(U0 × {t}) maps U0 to some other set. It is convenient to think of each point u being allowed to “flow for time t” along the integral curve cu —see Figure 4.1.2 and our opening motivation.

Figure 4.1.2. Picturing a flow.

This is a picture of the set U0 “flowing”. The system (U0 , a, F ) is called a local flow of X, or sometimes a flow box . The analogous situation on a manifold is given by the following definition. 4.1.11 Definition. Let M be a manifold and X a C r vector field on M , r ≥ 1. A flow box of X at a point m ∈ M is a triple (U0 , a, F ), where (i) U0 ⊂ M is open, m ∈ U0 , and a ∈ R, where a > 0 or a = +∞; (ii) F : U0 × Ia → M is of class C r , where Ia = ]−a, a[; (iii) for each u ∈ U0 , cu : Ia → M defined by cu (t) = F (u, t) is an integral curve of X at the point u; (iv) if Ft : U0 → M is defined by Ft (u) = F (u, t), then for t ∈ Ia , Ft (U0 ) is open, and Ft is a C r diffeomorphism onto its image. 4.1.12 Proposition. Suppose (U0 , a, F ) is a triple satisfying (i), (ii), and (iii) of Definition 4.1.11. Then for t, s and t + s ∈ Ia , Ft+s = Ft ◦ Fs = Fs ◦ Ft

and

F0 is the identity map,

whenever the compositions above are defined. Moreover, if Ut = Ft (U0 ) and Ut ∩ U0 6= ∅, then Ft |U−t ∩ U0 : U−t ∩ U0 → U0 ∩ Ut is a diffeomorphism and its inverse is F−t |U0 ∩ Ut . Proof. Ft+s (u) = cu (t + s), where cu is the integral curve defined by F at u. But d(t) = Ft (Fs (u)) = Ft (cu (s)) is the integral curve through cu (s) and f (t) = cu (t + s) is also an integral curve at cu (s). Hence by global uniqueness Proposition 4.1.10 we have Ft (Fs (u)) = cu (t + s) = Ft+s (u). To show that Ft+s = Fs ◦ Ft , observe that Ft+s = Fs+t = Fs ◦ Ft . Since cu (t) is a curve at u, cu (0) = u, so F0 is the identity. Finally, the last statement is a consequence of Ft ◦ F−t = F−t ◦ Ft = identity. Note, however, that Ft (U0 ) ∩ U0 = ∅ can occur. 

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219

4.1.13 Proposition (Existence and Uniqueness of Flow Boxes). Let X be a C r vector field on a manifold M . For each m ∈ M there is a flow box of X at m. Suppose (U0 , a, F ), (U00 , a0 , F 0 ) are two flow boxes at m ∈ M . Then F and F 0 are equal on (U0 ∩ U00 ) × (Ia ∩ Ia0 ). Proof. (Uniqueness). Again we emphasize that this does not follow at once from Theorem 4.1.5, since U0 and U00 need not be chart domains. However, for each point u ∈ U0 ∩ U00 we have F |{u} × I = F 0 |{u} × I, where I = Ia ∩ Ia0 . This follows from Proposition 4.1.10 and Definition 4.1.11(iii). Hence F = F 0 on the set (U0 ∩ U00 ) × I. (Existence). Let (U, ϕ) be a chart in M with m ∈ U . It is enough to establish the result in ϕ(U ) by means of the local representation. Thus let (U00 , a, F 0 ) be a flow box of X, the local representative of X, at ϕ(m) as given by Theorem 4.1.5, with U00 ⊂ U 0 = ϕ(U )

and F 0 (U00 × Ia ) ⊂ U 0 ,

U0 = ϕ−1 (U00 )

and let F : U0 × Ia → M ;

(u, t) 7→ ϕ−1 (F 0 (ϕ(u), t)).

Since F is continuous, there is a b ∈ ]0, a[ ⊂ R and V0 ⊂ U0 open, with m ∈ V0 , such that F (V0 × Ib ) ⊂ U0 . We contend that (V0 , b, F ) is a flow box at m (where F is understood as the restriction of F to V0 ×Ib ). Parts (i) and (ii) of Definition 4.1.11 follow by construction and (iii) is a consequence of the remarks following Definition 4.1.4 on the local representation. To prove (iv), note that for t ∈ Ib , Ft has a C r inverse, namely, F−t as Vt ∩ U0 = Vt . It follows that Ft (V0 ) is open. And, since Ft and F−t are both of class C r , Ft is a C r diffeomorphism.  As usual, there is an analogous result for time- (or parameter-) dependent vector fields. Straightening Out Theorem. The following result shows that near a point m satisfying X(m) 6= 0, the flow can be transformed by a change of variables so that the integral curves become straight lines. Naturally, the proof makes use of the inverse function theorm. 4.1.14 Theorem (Straightening Out Theorem). Let X be a vector field on a manifold M and suppose at m ∈ M , X(m) 6= 0. Then there is a local chart (U, ϕ) with m ∈ U such that (i) ϕ(U ) = V × I ⊂ G × R = E, V ⊂ G, open, and I = ]−a, a[ ⊂ R, a > 0; (ii) ϕ−1 |{v} × I : I → M is an integral curve of X at ϕ−1 (v, 0), for all v ∈ V ; (iii) the local representative of X is given by Xlr (y, t) = (y, t; 0, 1). Proof. Since the result is local, by taking any initial coordinate chart, it suffices to prove the result in E. We can arrange things so that we are working near 0 ∈ E and X(0) = (0, 1) ∈ E = G ⊕ R where G is a complement to the span of X(0). Letting (U0 , b, F ) be a flow box for X at 0 where U0 = V0 × ]−ε, ε[ and V0 is open in G, define f0 : V0 × Ib → E by f0 (y, t) = Ft (y, 0). But Df0 (0, 0) = Identity since

∂Ft (0, 0) = X(0) = (0, 1) ∂t t=0

and F0 = Identity.

By the inverse mapping theorem there are open neighborhoods V × Ia ⊂ V0 × Ib and U = f0 (V × Ia ) of (0, 0) such that f = f0 |V × Ia : V × Ia → U is a diffeomorphism. Then f −1 : U → V × Ia can serve as chart for (i). Notice that c = f |{y} × I : I → U is the integral curve of X through (y, 0) for all y ∈ V , thus proving (ii). Finally, the expression of the vector field X in this local chart given by f −1 is Df −1 (y, t) · X(f (y, t)) = Df −1 (c(t)) · c0 (t) = (f −1 ◦ c)0 (t) = (0, 1), since (f −1 ◦ c)(t) = (y, t), thus proving (iii). 

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In §4.3 we shall see that singular points, where the vector field vanishes, are of great interest in dynamics. The straightening out theorem does not claim anything about these points. Instead, one needs to appeal to more sophisticated normal form theorems; see Guckenheimer and Holmes [1983]. Global Flows. Now we turn our attention from local flows to global considerations. These ideas center on considering the flow of a vector field as a whole, extended as far as possible in the t-variable. 4.1.15 Definition. Given a manifold M and a vector field X on M , let DX ⊂ M × R be the set of (m, t) ∈ M × R such that there is an integral curve c : I → M of X at m with t ∈ I. The vector field X is called complete if DX = M × R. A point m ∈ M is called σ-complete, where σ = +, −, or ±, if DX ∩ ({m} × R) contains all (m, t) for t > 0, < 0, or t ∈ R, respectively. Let T + (m) (resp., T − (m)) denote the sup (resp., inf ) of the times of existence of the integral curves through m; T + (m) (resp., T − (m)) is called the positive (resp., negative) lifetime of m. Thus, X is complete iff each integral curve can be extended so that its domain becomes ]−∞, ∞[; that is, T + (m) = ∞ and T − (m) = −∞ for all m ∈ M . 4.1.16 Examples. A. For M = R2 , let X be the constant vector field, whose principal part is (0, 1). Then X is complete since the integral curve of X through (x, y) is t 7→ (x, y + t). B. On M = R2 \{0}, the same vector field is not complete since the integral curve of X through (0, −1) cannot be extended beyond t = 1; in fact as t → 1 this integral curve tends to the point (0, 0). Thus T + (0, −1) = 1, while T − (0, −1) = −∞. C. On R consider the vector field X(x) = 1 + x2 . This is not complete since the integral curve c with c(0) = 0 is c (t) = tan−1 t and thus it cannot be continuously extended beyond −π/2 and π/2; that is, T ± (0) = ±π/2.  The main global properties of flows are given in the next result. 4.1.17 Theorem.

Let M be a manifold and X ∈ Xr (M ), r ≥ 1. Then

(i) DX ⊃ M × {0}; (ii) DX is open in M × R; (iii) there is a unique C r mapping FX : DX → M such that the mapping t 7→ FX (m, t) is an integral curve at m for all m ∈ M ; (iv) for (m, t) ∈ DX , (FX (m, t), s) ∈ DX iff (m, t + s) ∈ DX ; in this case FX (m, t + s) = FX (FX (m, t), s). Proof. Part (i) is obvious from the definitions. To show (ii), we choose a point (m0 , t0 ) ∈ DX and assume for definiteness that t0 > 0. We must show that there is a neighborhood U of m0 and a positive δ such that U ×]t0 − δ, t0 + δ[⊂ DX . First consider the case in which m0 is an equilibrium point. We assume for simplicity that t0 > 0. Given t0 , we must show that there is a neighborhood U of m0 and δ > 0 such that for all points m ∈ U , the integral curve Ft (m) through m exists for time at least t0 + δ. In fact, we choose any δ, such as δ = 1. We claim that on a small enough neighborhoo of m0 , the integral curve will stay close to m0 ; moreover, it is enough to show this locally in a Banach space and with the assumption that in the relevant chart, m0 corresponds to the origin. However, we claim that this local result follows directly from equation (4.1.2). Indeed, we start with the ball Bb/2 (0) as constructed in Lemma 4.1.8. Now we choose the neighborhood U of zero to be the

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ball of radius r = be−K(t0 +1) /2. Then the estimate (4.1.2) becomes, for x ∈ U and for 0 ≤ t ≤  (where  is defined in Lemma 4.1.8), kFt (x)k ≤ eKt kxk ≤ eK(t0 +1) kxk ≤ b/2. Thus, Ft (x) remains in Bb/2 (0) and so the time of existence of the integral curve starting at F (x) is also at least . Thus, the integral curve starting at x exists for a time at least 2 and for t =  + τ , where 0 ≤ τ ≤ , Ft (x) = F (Fτ (x)) and estimate as follows: kF+τ (x)k = k (F ◦ Fτ ) (x)k ≤ eK kFτ (x)k ≤ eK eKτ kxk = eKt kxk ≤ eK(t0 +1) kxk ≤ b/2. This argument shows that the integral curve through x not only exists for a time at least 2, it remains in the ball Bb/2 (0). The argument now can be repeated until we reach the time at least t0 + 1 − . One can then proceed for time at least  on the final step. The above estimate is then not required for the final step. The second case is when m0 is not an equilibrium; by global uniqueness, at all points along the the integral curve c(t) with c(0) = m0 , the vector field X does not vanish. Around each point c(t¯) on the curve, choose a neighborhood U given by the Straightening Out Theorem 4.1.14. By compactness of the interval [0, t0 ], we can cover the curve with a finite number of such neighborhoods, say U0 , . . . , Un−1 , Un associated with points m0 = c(0), m1 = c(t(1)), . . . , mn−1 = c(tn−1 ), mn = c(t(n)) = c(t0 ). Each of these neighborhoods has a uniform time of existence which we denote by 2α(0), 2α(1), . . ., 2α(n − 1), 2α(n), so that as t ranges from [ti − α(i), ti + α(i)], the point c(t) ranges from one end of Ui to the other. Pick points pi ∈ Ui ∩ Ui+1 that also lie on the curve c(t). Let s(i) be such that c(s(i)) = pi , i = 1, . . . , n. See Figure 4.1.3.

U1 U0 m0

U2

m1 p1

Un-1

pn

Un

m2 p2

mn-1

Figure 4.1.3. A chain of charts.

By construction, these times s(i) are such that s(i) < s(i + 1). Because the vector field in local representation on the sets Ui has been straightened out, we also have s(i+1)−t(i) < α(i) and t(i+1)−s(i+1) < α(i+1). In addition, we similarly have s(i + 1) − s(i) < 2α(i). Let W1 = U0 ∩ U1 . Since s(2) − s(1) < 2α(1) and any integral curve starting in W1 ⊂ U1 exists for time at least 2α(1), the domain of Fs(2)−s(1) includes W1 and hence Fs(2)−s(1) (W1 ) is defined. Define the open set W2 = Fs(2)−s(1) (W1 ) ∩ U1 ∩ U2 and use s(3) − s(2) < 2α(2) and the fact that W2 ⊂ U2 to conclude that the domain of Fs(3)−s(2) contains U2 . Define the open set W3 = Fs(3)−s(2) (W2 ) ∩ U2 ∩ U3 and inductively define Wi = Fs(i)−s(i−1) (Wi−1 ) ∩ Ui−1 ∩ Ui ,

i = 1, . . . , n,

which are open sets. Since s(1) < α(0) and W1 ⊂ U0 , the domain of F−s(1) includes W1 and thus V1 = F−s(1) (W1 ) ∩ U0 is an open neighborhood of m0 . Since s(2) − s(1) < 2α(1) and W2 ⊂ U1 , the domain of F−s(2)+s(1) contains W2 and so F−s(2)+s(1) (W2 ) ⊂ W1 is defined. Therefore F−s(2) (W2 ) = F−s(1) (F−s(2)+s(1) (W2 )) ⊂ F−s(1) (W1 )

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W1

W2

U1

U2

V 1 V2 U0

m0

m2 m1 p1

p2

Figure 4.1.4. The construction used to prove the openness of DX .

exists and is an open neighborhood of m0 . Put V2 = F−s(2) (W2 ) ∩ U0 . Now proceed inductively to show that F−s(i) (Wi ) ⊂ F−s(i−1) (Wi−1 ) makes sense and is an open neighborhood of m0 , i = 1, 2, . . . , n; see Figure 4.1.4. Let Vi = F−s(i) (Wi ) ∩ U0 , i = 1, . . . , n, which are open neighborhoods containing m0 . Any integral curve b(t) starting in Vn exists thus for time at least s(n) and Fs(n) (Vn ) ⊂ Wn ⊂ Un . Now consider the integral curve starting at b(s(n)) whose time of existence is at least 2α(n). By uniqueness, b(t) can be smoothly extended to an integral curve which exists for time at least s(n) + 2α(n) > tn + α(n). Thus, we can choose δ = α(n) and the neighborhood of m0 that we sought can be taken to be Vn . To prove (iii), note that we get a unique map FX : DX → M by the global uniqueness and local existence of integral curves: (m, t) ∈ DX when the integral curve m(s) through m exists for s ∈ [0, t]. We set FX (m, t) = m(t). To show FX is C r , note that in a neighborhood of a fixed m0 and for small t, it is C r by local smoothness. To show FX is globally C r , first note that (iv) holds by global uniqueness. Then in a neighborhood of the compact set { m(s) | s ∈ [0, t] } we can write FX as a composition of finitely many C r maps by taking short enough time steps so the local flows are C r .  4.1.18 Definition. Let M be a manifold and X ∈ Xr (M ), r ≥ 1. Then the mapping FX is called the integral of X, and the curve t 7→ FX (m, t) is called the maximal integral curve of X at m. In case X is complete, FX is called the flow of X. Thus, if X is complete with flow F , then the set { Ft | t ∈ R } is a group of diffeomorphisms on M , sometimes called a one-parameter group of diffeomorphisms. Since Fn = (F1 )n (the nth power), the notation F t is sometimes convenient and is used where we use Ft . For incomplete flows, (iv) says that Ft ◦ Fs = Ft+s wherever it is defined. Note that Ft (m) is defined for t ∈ ]T − (m), T + (m)[. The reader should write out similar definitions for the time-dependent case and note that the lifetimes depend on the starting time t0 . 4.1.19 Proposition. Let X be C r , where r ≥ 1. Let c(t) be a maximal integral curve of X such that for every finite open interval ]a, b[ in the domain ]T − (c(0)), T + (c(0))[ of c, c(]a, b[) lies in a compact subset of M . Then c is defined for all t ∈ R. Proof. It suffices to show that a ∈ I, b ∈ I, where I is the interval of definition of c. Let tn ∈ ]a, b[, tn → b. By compactness we can assume some subsequence c(tn(k) ) converges, say, to a point x in M . Since the domain of the flow is open, it contains a neighborhood of (x, 0). Thus, there are ε > 0 and τ > 0 such that integral curves starting at points (such as c(tn(k) ) for large k) closer than ε to x persist for a time longer than τ . This serves to extend c to a time greater than b, so b ∈ I since c is maximal. Similarly, a ∈ I.  The support of a vector field X defined on a manifold M is defined to be the closure of the set { m ∈ M | X(m) 6= 0 }. 4.1.20 Corollary. A C r vector field with compact support on a manifold M is complete. In particular, a C r vector field on a compact manifold is complete.

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Completeness corresponds to well-defined dynamics persisting eternally. In some circumstances (shock waves in fluids and solids, singularities in general relativity, etc.) one has to live with incompleteness or overcome it in some other way. Because of its importance we give two additional criteria. In the first result we use the notation X[f ] = df · X for the derivative of f in the direction X. Here f : E → R and df stands for the derivative map. In standard coordinates on Rn ,  df (x) =

∂f ∂f ,..., n ∂x1 ∂x

 and

X[f ] =

n X

Xi

i=1

∂f . ∂xi

4.1.21 Proposition. Suppose X is a C k vector field on the Banach space E, k ≥ 1, and f : E → R is a C 1 proper map; that is, if {xn } is any sequence in E such that f (xn ) → a, then there is a convergent subsequence {xn(i) }. Suppose there are constants K, L ≥ 0 such that |X[f ](m)| ≤ K|f (m)| + L

for all m ∈ E.

Then the flow of X is complete. Proof.

From the chain rule we have (∂/∂t)f (Ft (m)) = X[f ](Ft (m)), so that t

Z f (Ft (m)) − f (m) =

X[f ](Fτ (m)) dτ 0

Applying the hypothesis and Gronwall’s inequality we see that |f (Ft (m))| is bounded and hence relatively compact on any finite t-interval, so as f is proper, a repetition of the argument in the proof of Proposition 4.1.19 applies.  Note that the same result holds if we replace “properness” by “inverse images of compact sets are bounded” and assume X has a uniform existence time on each bounded set. This version is useful in some infinite dimensional examples. 4.1.22 Proposition. Let X be a C r vector field on the Banach space E, r ≥ 1. Let σ be any integral curve of X. Assume kX(σ(t))k is bounded on finite t-intervals. Then σ(t) exists for all t ∈ R. Proof.

Suppose kX(σ(t))k < A for t ∈ ]a, b[ and let tn → b. For tn < tm we have Z

tm

kσ(tn ) − σ(tm )k ≤

0

Z

tm

kσ (t)k dt = tn

kX(σ(t))k dt ≤ A|tm − tn |. tn

Hence σ(tn ) is a Cauchy sequence and therefore, converges. Now argue as in Proposition 4.1.19.



4.1.23 Examples. A. Let X be a C r vector field, r ≥ 1, on the manifold M admitting a first integral , that is, a function f : M → R such that X[f ] = 0. If all level sets f −1 (r), r ∈ R are compact, X is complete. Indeed, each integral curve lies on a level set of f so that the result follows by Proposition 4.1.19. ¨ (t) = B. Newton’s equations for a moving particle of mass m in a potential field in Rn are given by q −(1/m)∇V (q(t)), for V : Rn → R a smooth function. We shall prove that if there are constants a, b ∈ R, b ≥ 0 such that (1/m)V (q) ≥ a − bkqk2 , then every solution exists for all time. To show this, rewrite the second order equations as a first order system q˙ = (1/m)p, p˙ = −∇V (q) and note that the energy E(q, p) = (1/2m)kpk2 + V (q) is a first integral. Thus, for any solution (q(t), p(t)) we have β = E(q(t), p(t)) = E(q(0), p(0)) ≥ V (q(0)). We can assume β > V (q(0)), that is, p(0) 6= 0, for if p(t) ≡ 0, then the conclusion

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is trivially satisfied; thus there exists a t0 for which p(t0 ) 6= 0 and by time translation we can assume that t0 = 0. Thus we have Z t ˙ kq(s)k ds kq(t)k ≤ kq(t) − q(0)k + kq(0)k ≤ kq(0)k + 0 s   Z t 1 = kq(0)k + 2 β − V (q(s)) ds m 0 Z tp ≤ kq(0)k + 2(β − a + bkq(s)k2 ) ds 0

or in differential form

p d kq(t)k ≤ 2(β − a + bkq(t)k2 ) dt

whence

k q(t)k

Z t≤

k q(0)k

du p 2(β − a + bu2 )

(4.1.3)

Now let r(t) be the solution of the differential equation d2 r(t) d = − (a − br2 )(t) = 2br(t), dt2 dr which, as a second order equation with constant coefficients, has solutions for all time for any initial conditions. Choose 2 r(0) = kq(0)k, [r(0)] ˙ = 2(β − a + bkq(0)k2 ) and let r(t) be the corresponding solution. Since   d 1 r(t) ˙ 2 + a − br(t)2 = 0, dt 2 it follows that (1/2)r(t) ˙ 2 + a − br(t)2 = (1/2)r(0) ˙ 2 + a − br(0)2 = β, that is, dr(t) p = 2(β − a + br(t)2 ) dt whence Z

r(t)

t= k q(0)k

du p 2(β − α + βu2

(4.1.4)

Comparing the two expressions (4.1.3) and (4.1.4) and taking into account that the integrand is > 0, it follows that for any finite time interval for which q(t) is defined, we have kq(t)k ≤ r(t), that is, q(t) remains ˙ in a compact set for finite t-intervals. But then q(t) also lies in a compact set since ˙ kq(t)k ≤ 2(β − a + bkq(s)k2 ). Thus by Proposition 4.1.19, the solution curve (q(t), p(t)) is defined for any t ≥ 0. However, since (q(−t), p(−t)) is the value at t of the integral curve with initial conditions (−q(0), −p(0)), it follows that the solution also exists for all t ≤ 0. (This example will be generalized in §8.1 to any Lagrangian system on a complete Riemannian manifold whose energy function is kinetic energy of the metric plus a potential, with the potential obeying an inequality of the sort here). The following counterexample shows that the condition V (q) ≥ a−bkqk2 cannot be relaxed much further. Take n = 1 and V (q) = − 81 ε2 q 2+(4/ε) , ε > 0. Then the equation q¨ = ε(ε + 2)q 1+(4/ε) /4 has the solution q(t) = 1/(t − 1)ε/2 , which cannot be extended beyond t = 1.

4.1 Vector Fields and Flows

C.

225

Let E be a Banach space. Suppose X(x) = A · x + B(x),

where A is a bounded linear operator of E to E and B is sublinear ; that is, B : E → E is C r with r ≥ 1 and satisfies kB(x)k ≤ Kkxk + L for constants K and L. We shall show that X is complete by using Proposition 4.1.22. (In Rn , Proposition 4.1.21 can also be used with f (x) = kxk2 .) Let x(t) be an integral curve of X on the bounded interval [0, T ]. Then Z t x(t) = x(0) + (A · x(s) + B(x(s))) ds 0

Therefore, Z kx(t)k ≤ kx(0)k +

t

(kAk + K)kx(s)k ds + Lt. 0

By Gronwall’s inequality, kx(t)k ≤ (LT + kx(0)k)e(kAk+K)t . Hence, x(t) and so X(x(t)) remain bounded on bounded t-intervals.



Semicontinuity of Lifetimes. The next result, which technically is closely related to Theorem 4.1.17 shows that lifetimes cannot suddenly decrease (the student should construct an example to show that they can suddenly increase). 4.1.24 Proposition. Let X be a C r vector field on the manifold M , r ≥ 1, m0 ∈ M , and let T + (m0 ) (resp. T − (m0 )) be the positive (resp. negative) lifetime of m0 . Then for each ε > 0, there exists a neighborhood V of m0 such that for all m ∈ V , T + (m) > T + (m0 ) − ε (respectively, T − (m0 ) < T − (m0 ) + ε). [One says that T + (m0 ) is a lower semi-continuous function of m.] Proof. Cover the segment { Ft (m0 ) | t ∈ [0, T + (m0 ) − ε] } with a finite number of neighborhoods U0 , . . . , Un , each in a chart domain and such that ϕi (Ui ) is diffeomorphic to the open ball Bb(i)/2 (0) in E given in the proof of Lemma 4.1.8, where ϕi is the chart map. Let mi ∈ Ui be such that ϕi (mi ) = 0 and t(i) such that Ft(i) (m0 ) = mi , i = 0, . . . , n, t(0) = 0, t(n) = T + (m0 ) − ε. By Lemma 4.1.8 the time of existence of all integral curves starting in Ui is uniformly at least α(i) > 0. As we did in Figure 4.1.3, pick points pi ∈ Ui ∩ Ui+1 and let s(i) be such that Fs(i) (m0 ) = pi , i = 0, . . . , n − 1, s(0) = 0, p0 = m0 , s(i) < s(i + 1), s(i + 1) − t(i) < α(i), t(i + 1) − s(i + 1) < α(i + 1), s(i + 1) − s(i) < min(α(i), α(i + 1)). Let W1 = U0 ∩ U1 . Since s(2) − s(1) < α(1) and any integral curve starting in W1 ⊂ U1 exists for time at least α(1), the domain of Fs(2)−s(1) includes W1 and hence Fs(2)−s(1) (W1 ) makes sense. Define the open set W2 = Fs(2)−s(1) (W1 ) ∩ U1 ∩ U2 and use s(3) − s(2) < α(2), W2 ⊂ U2 to conclude that the domain of Fs(3)−s(2) contains U2 . Define the open set W3 = Fs(3)−s(2) (W2 ) ∩ U2 ∩ U3 and inductively define Wi = Fs(i)−s(i−1) (Wi − 1) ∩ Ui−1 ∩ Ui ,

i = 1, . . . , n,

which are open sets. Since s(1) < α(0) and W1 ⊂ U0 , the domain of F−s(1) includes W1 and thus V1 = F−s(1) (W1 ) ∩ U0 is an open neighborhood of m0 . Since s(2) − s(1) < α(1) and W2 ⊂ U1 , the domain of F−s(2)+s(1) contains W2 and so F−s(2)+s(1) (W2 ) ⊂ W1 makes sense. Therefore F−s(2) (W2 ) = F−s(1) (F−s(2)+s(1) (W2 )) ⊂ F−s(1) (W1 ) exists and is an open neighborhood of m0 . Put V2 = F−s(2) (W2 ) ∩ U0 . Now proceed inductively to show that F−s(i) (Wi ) ⊂ F−s(i−1) (Wi−1 ) makes sense and is an open neighborhood of m0 , i = 1, 2, . . . , n, as we did in

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Figure 4.1.4. Let Vi = F−s(i) (Wi ) ∩ U0 , i = 1, . . . , n, open neighborhoods containing m0 . Any integral curve c(t) starting in Vn exists thus for time at least s(n) and Fs(n) (Vn ) ⊂ Wn ⊂ Un . Now consider the integral curve starting at c(s(n)) whose time of existence is at least α(n). By uniqueness, c(t) can be smoothly extended to an integral curve which exists for time at least s(n) + α(n) > tn = T + (m0 ) − ε.  The same result and proof hold for time dependent vector fields depending on a parameter. 4.1.25 Corollary. Let Xt be a C r time-dependent vector field on M , r ≥ 1, and let m0 be an equilibrium of Xt , that is, Xt (m0 ) = 0 for all t. Then for any T there exists a neighborhood V of m0 such that any m ∈ V has integral curve existing for time t ∈ [−T, T ]. Proof.

Since T + (m0 ) = +∞, T − (m0 ) = −∞, the previous proposition gives the result.



Supplement 4.1A Product Formulas A result of some importance in both theoretical and numerical work concerns writing a flow in terms of iterates of a known mapping. Let X ∈ X(M ) with flow Ft (maximally extended). Let Kε (x) be a given map defined in some open set of [0, ∞[ × M containing {0} × M and taking values in M , and assume that (i) K0 (x) = x and (ii) Kε (x) is C 1 in ε with derivative continuous in (ε, x). We call K the algorithm. 4.1.26 Theorem. Let X be a C r vector field, r ≥ 1. Assume that the algorithm Kε (x) is consistent with X in the sense that ∂ X(x) = Kε (x) ∂ε ε=0 Then, if (t, x) is in the domain of Ft (x), (Kt/n )n (x) is defined for n sufficiently large and converges to Ft (x) as n → ∞. Conversely, if (Kt/n )n (x) is defined and converges for 0 ≤ t ≤ T , then (T, x) is in the domain of F and the limit is Ft (x). In the following proof the notation O(x), x ∈ R is used for any continuous function in a neighborhood of the origin such that O(x)/x is bounded. Recall from §2.1 that o(x) denotes a continuous function in a neighborhood of the origin satisfying limx→0 o(x)/x = 0. Proof. First, we prove that convergence holds locally. We begin by showing that for any x0 , the iterates (Kt/n )n (x0 ) are defined if t is sufficiently small. Indeed, on a neighborhood of x0 , Kε (x) = x + O(ε), so if (Kt/j )j (x) is defined for x in a neighborhood of x0 , for j = 1, . . . , n − 1, then (Kt/n )n (x) − x = ((Kt/n )n x − (Kt/n )n−1 x) + ((Kt/n )n−1 − (Kt/n )n−2 x) + · · · + (Kt/n (x) − x) = O(t/n) + · · · + O(t/n) = O(t). This is small, independent of n for t sufficiently small; so, inductively, (Kt/n )n (x) is defined and remains in a neighborhood of x0 for x near x0 .

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227

Let β be a local Lipschitz constant for X so that kFt (x) − Ft (y)k ≤ eβ|t| kx − yk. Now write Ft (x) − (Kt/n )n (x) = (Ft/n )n (x) − (Kt/n )n (x) = (Ft/n )n−1 Ft/n (x) − (Ft/n )n−1 Kt/n (x) +(Ft/n )n−2 Ft/n (y1 ) − (Ft/n )n−2 Kt/n (y1 ) + · · · + (Ft/n )n−k Ft/n (yk−1 ) − (Ft/n )n−k Kt/n (Yk−1 ) + · · · + Ft/n (yn−1 ) − Kt/n (yn−1 ) where yk = (Kt/n )k (x). Thus n

kFt (x) − (Kt/n ) (x)k ≤

n X

eβ|t|(n−k)/n kFt/n (tk−1 ) − Kt/n (yk−1 )k

k=1

≤ neβ|t| o(t/n) → 0

as

n → ∞,

since Fε (y) − Kε (y) = o(ε) by the consistency hypothesis. Now suppose Ft (x) is defined for 0 ≤ t ≤ T . We shall show (Kt/n )n (x) converges to Ft (x). By the foregoing proof and compactness, if N is large enough, Ft/N = limn→∞ (Kt/nN )n uniformly on a neighborhood of the curve t 7→ Ft (x). Thus, for 0 ≤ t ≤ T , Ft (x) = (Ft/N )N (x) = lim (Kt/nN )N (x). n→∞

By uniformity in t, FT (x) = lim (KT /j )j (x). j→∞

n

Conversely, let (Kt/n ) (x) converge to a curve c(t), 0 ≤ t ≤ T . Let S = { t | Ft (x) is defined and c(t) = Ft (x) }. From the local result, S is a nonempty open set. Let t(k) ∈ S, t(k) → t. Thus Ft(k) (x) converges to c(t), so by local existence theory, Ft (x) is defined, and by continuity, Ft (x) = c(t). Hence S = [0, T ] and the proof is complete.  4.1.27 Corollary.

Let X, Y ∈ X(M ) with flows Ft and Gt . Let St be the flow of X + Y . Then for x ∈ M , St (x) = lim (Ft/n ◦ Gt/n )n (x). n→∞

The left-hand side is defined iff the right-hand side is. This follows from Theorem 4.1.26 by setting Kε (x) = (Fε ◦ Gε )(x). For example, for n × n matrices A and B, Corollary 4.1.27 yields the classical formula e(A+B) = lim (eA/n eB/n )n . n→∞

To see this, define for any n × n matrix C a vector field XC ∈ X(Rn ) by XC (x) = Cx. Since XC is linear in C and has flow Ft (x) = etC x, the formula follows from Corollary 4.1.27 by letting t = 1. The topic of this supplement will continue in Supplement 4.2A. The foregoing proofs were inspired by Nelson [1969] and Chorin, Hughes, McCracken, and Marsden [1978].

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Supplement 4.1B Invariant Sets If X is a smooth vector field on a manifold M and N ⊂ M is a submanifold, the flow of X will leave N invariant (as a set) iff X is tangent to N . If N is not a submanifold (e.g., N is an open subset together with a non-smooth boundary) the situation is not so simple; however, for this there is a nice criterion going back to Nagumo [1942]. Our proof follows Brezis [1970]. 4.1.28 Theorem. Let X be a locally Lipschitz vector field on an open set U ⊂ E, where E is a Banach space. Let G ⊂ U be relatively closed and set d(x, G) = inf{ kx − yk | y ∈ G }. The following are equivalent: (i) limh↓0 (d(x + hX(x), G)/h) = 0 locally uniformly in x ∈ G (or pointwise if E = Rn ); (ii) if x(t) is the integral curve of X starting in G, then x(t) ∈ G for all t ≥ 0 in the domain of x(·). Note that x(t) need not lie in G for t ≤ 0; so G is only + invariant. (We remark that if X is only continuous the theorem fails.) We give the proof assuming E = Rn for simplicity. Proof.

Assume (ii) holds. Setting x(t) = Ft (x), where Ft is the flow of X and x ∈ G, for small h we get



x(h) − x

, − X(x) d(x + hX(x), G) ≤ kx(h) − x − hX(x)k = |h|

h

from which (i) follows. Now assume (i). It suffices to show x(t) ∈ G for small t. Near x = x(0) ∈ G, say on a ball of radius r, we have kX(x1 ) − X(x2 )k ≤ Kkx1 − x2 k and kFt (x1 ) − Ft (x2 )k ≤ eKt kx1 − x2 k. We can assume kFt (x) − xk < r/2. Set ϕ(t) = d(Ft (x), G) and note that ϕ(0) = 0, so that for small t, ϕ(t) < r/2. Since G is relatively closed, and E = Rn , d(Ft (x), G) = kFt (x) − yt k for some yt ∈ G. (In the general Banach space case an approximation argument is needed here.) Thus, kyt − xk < r. For small h, kFh (yt ) − xk < r, so that ϕ(t + h) = inf kFt+h (x) − zk z∈G

≤ inf {kFt+h (x) − Fh (yt )k + kFh (yt ) − yt − hX(yt )k z∈G

+kyt + hX(yt ) − zk} = kFt+h (x) − Fh (yt )k + kFh (yt ) − yt − hX(yt )k +d(yt + hX(yt ), G) ≤ eKh kyt − Ft (x)k + kFh (yt ) − yt − hX(yt )k +d(yt + hX(yt ), G) or ϕ(t + h) − ϕ(t) ≤ h



Fh (yt ) − yt

eKh − 1

ϕ(t) + − X(yt )

h h 1 + d(yt + hX(yt ), G). h 

4.1 Vector Fields and Flows

Hence lim sup h↓0

229

ϕ(t + h) − ϕ(t) ≤ Kϕ(t). h

As in Gronwall’s inequality, we may conclude that ϕ(t) ≤ eKt ϕ(0), so ϕ(t) = 0.



4.1.29 Example. Let X be a C ∞ vector field on Rn , let g : Rn → R be smooth, and let λ ∈ R be a regular value for g, so that g −1 (λ) is a submanifold. Refer to Figure 4.1.5. Let G = g −1 (]−∞, λ]) and suppose that on g −1 (λ), hX, grad gi ≤ 0.

x

Figure 4.1.5. The set G = g −1 (] − ∞, λ[) is invariant if X does not point strictly outwards at ∂G.

Then G is + invariant under Ft as may be seen by using Theorem 4.1.28. This result has been generalized to the case where ∂G might not be smooth by Bony [1969]. See also Redheffer [1972] and Martin [1973]. Related references are Yorke [1967], Hartman [1972], and Crandall [1972]. 

Supplement 4.1C A second Proof of the Existence and Uniqueness of Flow Boxes We now give an alternative “modern” proof of Theorem 4.1.5 and Proposition 4.1.13, namely, if X ∈ Xk (M ), k ≥ 1, then for each m ∈ M there exists a unique C k flow box at m. The basic idea is due to Robbin [1968] although similar alternative proofs were simultaneously discovered by Abraham and Pugh [unpublished] and Marsden [1968b, p. 368]. The present exposition follows Robbin [1968] and Ebin and Marsden [1970]. Step 1

Existence and uniqueness of integral curves for C 1 vector fields.

Proof. Working in a local chart, we may assume that X : Dr (0) → E, where Dr (0) is the open disk at the origin of radius r in the Banach space E. Let U = Dr/2 (0), I = [−1, 1] and define Φ : R × C01 (I, U ) → C 0 (I, E) by Φ(s, γ)(t) =

dγ (t) − sX(γ(t)), dt

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where C i (I, E) is the Banach space of C i -maps of I into E, endowed with the k · ki -norm (see Supplement 2.4B), C0i (I, E) = { f ∈ C i (I, E) | f (0) = 0 } is a closed subspace of C i (I, E) and C0i (I, U ) = { f ∈ C0i (I, E) | f (I) ⊂ U } is open in C0i (I, E). We first show that Φ is a C 1 -map. The map d/dt : C01 (I, E) → C 0 (I, E) is clearly linear and is continuous since kd/dtk ≤ 1. Moreover, if dγ/dt = 0 on I, then γ is constant and since γ(0) = 0, it follows γ = 0; that is, d/dt is injective. Given δ ∈ C 0 (I, E), Z t γ(t) = δ(s) ds 0

defines an element of C0i (I, E) with dγ/dt = δ, that is, d/dt is a Banach space isomorphism from C01 (I, E) to C 0 (I, E). From these remarks and the Ω lemma 2.4.21, it follows that Φ is a C 1 -map. Moreover, Dγ Φ(0, 0) = d/dt is an isomorphism of C01 (I, E) with C 0 (I, E). Since Φ(0, 0) = 0, by the implicit function theorem there is an ε > 0 such that Φ(ε, γ) = 0 has a unique solution γε (t) in C01 (I, U ). The unique integral curve sought is γ(t) = γε (t/ε), −ε ≤ t ≤ ε.  The same argument also works in the time-dependent case. It also shows that γ varies continuously with X. Step 2.

The local flow of a C k vector field X is C k .

Proof. First, suppose k = 1. Modify the definition of Φ in Step 1 by setting Ψ : R × U × C01 (I, U ) → C 0 (I, E), Ψ(s, x, γ)(t) = γ 0 (t) − sX(x + γ(t)). As in Step 1, Ψ is a C 1 -map and Dγ Ψ(0, 0, 0) is an isomorphism, so Ψ(ε, x, γ) = 0 can be locally solved for γ giving a map Hε : U → C01 (I, U ), ε > 0. The local flow is F (x, t) = x + Hε (x)(t/ε), as in Step 1. By Proposition 2.4.20 (differentiability of the evaluation map), F is C 1 . Moreover, if v ∈ E, we have DFt (x) · v = v + (DHε (x) · v)(t/ε), so that the mixed partial derivative   1 t d DFt (x) · v = (DHε (x) · v) dt ε ε exists and is jointly continuous in (t, x). By Exercise 2.4-7, D(dFt (x)/dt) exists and equals (d/dt)(DFt (x)). Next we prove the result for k ≥ 2. Consider the Banach space F = C k−1 (cl(U ), E) and the map ωX : F → F ; η 7→ X ◦ η. This map is C 1 by the Ω lemma (remarks following Lemma 2.4.21). Regarding ωX as a vector field on F , it has a unique C 1 integral curve ηt with η0 = identity, by Step 1. This integral curve is the local flow of X and is C k−1 since it lies in F . Since k ≥ 2, ηt is at least C 1 and so one sees that Dηt = ut satisfies dut /dt = DX(ηt ) · ut , so by Step 1 again, ut lies in C k−1 . Hence ηt is C k . The following is a useful alternative argument for proving the result for k = 1 from that for k ≥ 2. For k = 1, let X n → X in C 1 , where X n are C 2 . By the above, the flows of X n are C 2 and by Step 1, converge uniformly that is, in C 0 , to the flow of X. From the equations for Dηtn , we likewise see that Dηt n converges uniformly to the solution of dut /dt = DX(ηt ) · ut , u0 = identity. It follows by elementary analysis (see Exercise 2.4-10 or Marsden and Hoffman [1993, p. 109]) that ηt is C 1 and Dηt = ut . 

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231

This proof works with minor modifications on manifolds with vector fields and flows of Sobolev class H s or Holder class C k+α ; see Ebin and Marsden [1970] and Bourguignon and Brezis [1974]. In fact the foregoing proof works in any function spaces for which the Ω lemma can be proved. Abstract axioms guaranteeing this are given in Palais [1968].

Exercises  4.1-1. Find an explicit formula for the flow Ft : R2 → R2 of the harmonic oscillator equation x ¨ + ω 2 x = 0, ω ∈ R a constant.  4.1-2. Show that if (U0 , a, F ) is a flow box for X, then (U0 , a, F− ) is a flow box for −X, where F− (u, t) = F (u, −t) and (−X)(m) = −(X(m)).  4.1-3. Show that the integral curves of a C r vector field X on an n-manifold can be defined locally in the neighborhood of a point where X is nonzero by n equations ψi (m, t) = ci = constant, i = 1, . . . , n in the n + 1 unknowns (m, t). Such a system of equations is called a local complete system of integrals. Hint: Use the straightening-out theorem.  4.1-4.

Prove the following generalization of Gronwall’s inequality. Suppose that v(t) ≥ 0 satisfies Z t v(t) ≤ C + p(s) v(s) ds , 0

1

where C ≥ 0 and p ∈ L . Then Z v(t) ≤ C exp

t

 |p(s)| ds .

0

Use this to generalize Example 4.1.23C to allow A to be a time-dependent matrix.  4.1-5.

Let Ft = etX be the flow of a linear vector field X on E. Show that the solution of the equation x˙ = X(x) + f (x)

with initial conditions x0 satisfies the variation of constants formula Z t tX x(t) = e x0 + e(t−s)X f (x(s)) ds 0 ∞

 4.1-6. Let F (m, t) be a C mapping of M × R to M such that Ft+s = Ft ◦ Fs and F0 = identity (where Ft (m) = F (m, t)). Show that there is a unique C ∞ vector field X whose flow is F .  4.1-7. Let σ(t) be an integral curve of a vector field X and let g : M → R. Let τ (t) satisfy τ 0 (t) = g(σ(τ (t))). Then show t 7→ σ(τ (t)) is an integral curve of gX. Show by example that even if X is complete, gX need not be.  4.1-8. (i) (Gradient Flows.) Let f : Rn → R be C 1 and let X = (∂f /∂x1 , . . . , ∂f /∂xn ) be the gradient of f . Let F be the flow of X. Show that f (Ft (x) ≥ f (Fs (x)) if t ≥ s. (ii) Use (i) to find a vector field X on Rn such that X(0) = 0, X 0 (0) = 0, yet 0 is globally attracting; that is, every integral curve converges to 0 as t → ∞. This exercise continues in Exercise 4.3-11.  4.1-9. Let c be a locally Lipschitz increasing function, c(t) > 0 for t ≥ 0 and assume that the differential equation r0 (t) = c(r(t)) has the solution with r(0) = r0 ≥ 0 existing for time t ∈ [0, T ]. Conclude that r(t) ≥ 0 for t ∈ [0, T ]. Prove the following comparison lemmas.

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(i) If h(t) is a continuous function on [0, T ], h(t) ≥ 0, satisfying h0 (t) ≤ c(h(t)) on [0, T ], h(0) = r0 , then show that h(t) ≤ r(t). Hint: Prove that h(t)

Z

r0

Z

dx ≤t= c(x)

r(t)

r0

dx c(x)

and use strict positivity of the integrand. (ii) Generalize (i) to the case h(0) ≤ r0 . Hint: The function h(t) = h(t) + r0 − h(0) ≤ h(t) satisfies the hypotheses in (i). (iii) If f (t) is a continuous function on [0, T ], f (t) ≥ 0, satisfying 1

Z f (t) ≤ r0 +

c(f (s)) ds 0

on [0, T ], then show that f (t) ≤ r(t). Hint: Z

1

c(f (s)) ds ≥ f (t)

h(t) = r0 + 0

satisfies the hypothesis in (i) since h0 (t) = c(f (t)) ≤ c(h(t)). (iv) If in addition Z 0

∞

dx = +∞ c(x)

show that the solution f (t) ≥ 0 exists for all t ≥ 0. (v) If h(t) is only continuous on [0, T ] and Z h(t) ≤ r0 +

t

c(h(s)) ds, 0

show that h(t) ≤ r(t) on [0, T ]. Hint: Approximate h(t) by a C 1 -function g(t) and show that g(t) satisfies the same inequality as h(t).  4.1-10. (i) Let X = y 2 ∂/∂x and Y = x2 ∂/∂y. Show that X and Y are complete on R2 but X + Y is not. Hint: Note that x3 − y 3 = constant and consider an integral curve with x(0) = y(0). (ii) Prove the following theorem: Let H be a Hilbert space and X and Y be locally Lipschitz vector fields that satisfy the following: (a) X and Y are bounded and Lipschitz on bounded sets; (b) there is a constant β ≥ 0 such that hY (x), xi ≤ βkxk2

for all x ∈ H;

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233

(c) there is a locally Lipschitz monotone increasing function c(t) > 0, t ≥ 0, such that Z ∞ dx = +∞ c(x) 0 and if x(t) is an integral curve of X, d kx(t)k ≤ c(kx(t)k). dt Then X, Y and X + Y are positively complete. Note: One may assume kX(x0 )k ≤ c(kx0 k) in (c) instead of (d/dt)kx(t)k ≤ c(kx(t)k). Hint: Find a differential inequality for (1/2)(d/dt)ku(t)k2 , where u(t) is an integral curve of X + Y and then use Exercise 4.1-9iii.  4.1-11. Prove the following result on the convergence of flows: Let Xα be locally Lipschitz vector fields on M for α in some topological space. Suppose the Lipschitz constants of Xα are locally bounded as α → α(0) and Xα → Xα(0) locally uniformly. Let c(t) be an integral curve of Xα(0) , 0 ≤ t ≤ T and ε > 0. Then the integral curves cα (t) of Xα with cα(0) = c(0) are defined for the interval t ∈ [0, T − ε] for α sufficiently close to α(0) and cα (t) → c(t) uniformly in t ∈ [0, T − ε] as α → α(0). If the flows are complete Ftα → Ft locally uniformly. (The vector fields may be time dependent if the estimates are locally t-uniform.) Hint: Show that Z t kcα (t) − c(t)k ≤ k kcα (τ ) − c(τ )k dτ 0 Z t + kXα (c(τ )) − Xα(0) (c(τ ))k dτ 0

and conclude from Gronwall’s inequality that cα (t) → c(t) for α → α(0) since the second term → 0. This estimate shows that cα (t) exists as long as c(t) does on any compact subinterval of [0, T [.  4.1-12. Prove that the C r flow of a C r+1 vector field is a C 1 function of the vector field by utilizing Supplement 4.1C. (Caution. It is known that the C k flow of a C k vector field cannot be a C 1 function of the vector field; see Ebin and Marsden [1970] for the explanation and further references).  4.1-13 (Nonunique integral curves on non-Hausdorff manifolds). Let M be the line with two origins (see Exercise 3.5-8) and consider the vector field X : M → T M which is defined by X([x, i]) = x, i = 1, 2; here [x, i] denotes a point of the quotient manifold M . Show that through every point other than [0, 0] and [0, 1], there are exactly two integral curves of X. Show that X is complete. Hint: The two distinct integral curves pass respectively through [0, 0] and [0, 1].  4.1-14.

Give another proof of Theorem 4.1.5 using Exercise 2.5-10.

 4.1-15.

Give examples of vector fields satisfying the following conditions:

(i) on R and S 1 with no critical points; generalize to Rn and Tn ; (ii) on S 1 with exactly k critical points; generalize to Tn ; (iii) on R2 and RP2 with exactly one critical point and all other orbits closed; (iv) on the M¨ obius band with no critical points and such that the only integral curve intersecting the zero section is the zero section itself;

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4. Vector Fields and Dynamical Systems

(v) on S 2 with precisely two critical points and one closed orbit; (vi) on S 2 with precisely one critical point and no closed orbit; (vii) on S 2 with no critical points on a great circle and nowhere tangent to it; show that any such vector field has its integral curves intersecting this great circle at most once; (viii) on T2 with no critical points, all orbits closed and winding exactly k times around T2 .  4.1-16. Let π : M → N be a surjective submersion. A vector field X ∈ X(M ) is called π-vertical if T π ◦ X = 0. If K is the Klein bottle, show that π : K → S 1 given by π([a, b]) = e2πia is a surjective submersion; see Figure 4.1.6. Prove that K is a non trivial S 1 -bundle.

6 K -? π ?

- S1

Figure 4.1.6. The Klein bottle as an S 1 -bundle

Hint: If it were trivial, there would exist a nowhere zero vertical vector field on K. In Figure 4.1.6, this means that arrows go up on the left and down on the right hand side. Follow a path from left to right and argue by the intermediate value theorem that the vector field must vanish somewhere.

4.2

Vector Fields as Differential Operators

In the previous section vector fields were studied from the point of view of dynamics; that is in terms of the flows they generate. Before continuing the development of dynamics, we shall treat some of the algebraic aspects of vector fields. The specific goal of the section is the development of the Lie derivative of functions and vector fields and its relationship with flows. One important feature is the behavior of the constructions under mappings. The operations should be as natural or covariant as possible when subjected to a mapping. Push-forward and Pull-Back of Functions and Vector Fields. We begin with a discussion of the action of mappings on functions and vector fields. First, recall some notation. Let C r (M, F) denote the space of C r maps f : M → F, where F is a Banach space, and let Xr (M ) denote the space of C r vector fields on M . Both are vector spaces with the obvious operations of addition and scalar multiplication. For brevity we write F(M ) = C ∞ (M, R), F r (M ) = C r (M, R) and X(M ) = X∞ (M ). Note that F r (M ) has an algebra structure; that is, for f, g ∈ F r (M ) the product f g defined by (f g)(m) = f (m)g(m) obeys the usual algebraic properties of a product such as f g = gf and f (g + h) = f g + f h. 4.2.1 Definition. (i) Let ϕ : M → N be a C r mapping of manifolds and f ∈ F r (N ). Define the pull-back of f by ϕ by ϕ∗ f = f ◦ ϕ ∈ F r (M ).

4.2 Vector Fields as Differential Operators

235

(ii) If ϕ is a C r diffeomorphism and X ∈ Xr (M ), the push-forward of X by ϕ is defined by ϕ∗ X = T ϕ ◦ X ◦ ϕ−1 ∈ Xr (N ). Consider local charts (U, χ), χ : U → U 0 ⊂ E on M and (V, ψ), ψ : V → V 0 ⊂ F on N , and let (T χ ◦ X ◦ χ−1 )(u) = (u, X(u)), where X : U 0 → E is the local representative of X. Then from the chain rule and the definition of push–forward, the local representative of ϕ∗ X is (T ψ ◦ (ϕ∗ X) ◦ ψ −1 )(v) = (v, D(ψ ◦ ϕ ◦ χ−1 )(u) · X(u)), where v = (ψ ◦ ϕ ◦ χ−1 )(u). The different point of evaluation on each side of the equation corresponds to the necessity of having ϕ−1 in the definition. If M and N are finite dimensional, xi are local coordinates on M and y j local coordinates on N , the preceding formula gives the components of ϕ∗ X by (ϕ∗ X)j (y) =

∂ϕj (x) X i (x) ∂xi

where y = ϕ(x). We can interchange “pull-back” and “push-forward” by changing ϕ to ϕ−1 , that is, defining ϕ∗ (resp. ϕ∗ ) by ϕ∗ = (ϕ−1 )∗ (resp. ϕ∗ = (ϕ−1 )∗ ). Thus the push-forward of a function f on M is ϕ∗ f = f ◦ ϕ−1 and the pull-back of a vector field Y on N is ϕ∗ Y = (T ϕ)−1 ◦ Y ◦ ϕ (Figure 4.2.1). Notice that ϕ must be a diffeomorphism in order that the pull-back and push-forward operations make sense, the only exception being pull-back of functions. Thus vector fields can only be pulled back and pushed forward by diffeomorphisms. However, even when ϕ is not a diffeomorphism we can talk about ϕ-related vector fields as follows.

objects defined on M

M

objects defined on N

N

Figure 4.2.1. Push–forward and pull–back.

4.2.2 Definition. Let ϕ : M → N be a C r mapping of manifolds. The vector fields X ∈ Xr−1 (M ) and Y ∈ Xr−1 (N ) are called ϕ-related , denoted X ∼ϕ Y , if T ϕ ◦ X = Y ◦ ϕ. Note that if ϕ is diffeomorphism and X and Y are ϕ-related, then Y = ϕ∗ X. In general however, X can be ϕ-related to more than one vector field on N . ϕ-relatedness means that the following diagram commutes:

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4. Vector Fields and Dynamical Systems

Tϕ

TM 6 X M

- TN 6 Y - N

ϕ

4.2.3 Proposition. (i) Pull-back and push-forward are linear maps, and ϕ∗ (f g) = (ϕ∗ f )(ϕ∗ g),

ϕ∗ (f g) = (ϕ∗ f )(ϕ∗ g).

(ii) If Xi ∼ϕ Yi , i = 1, 2, and a, b ∈ R, then aX1 + bX2 ∼ϕ aY1 + bY2 . (iii) For ϕ : M → N and ψ : N → P , we have (ψ ◦ ϕ)∗ = ϕ∗ ◦ ψ ∗

and

(ψ ◦ ϕ)∗ = ψ∗ ◦ ϕ∗ .

(iv) If X ∈ X(M ), Y ∈ X(N ), Z ∈ X(P ), X ∼ϕ Y , and Y ∼ψ Z, then X ∼ψ◦ϕ Z. In this proposition it is understood that all maps are diffeomorphisms with the exception of the pull-back of functions and the relatedness of vector fields. Proof. The proof of (i) and (ii) consists of straightforward verifications. For example, if Xi ∼ϕ Yi , i = 1, 2, then T ϕ ◦ (aX1 + bX2 ) = aT ϕ ◦ X1 + bT ϕ ◦ X2 = aY1 ◦ ϕ + bY2 ◦ ϕ, that is, aX1 + bX2 ∼ϕ aY1 + bY2 . The relations in (iii) on functions are simple consequences of the definition, and on X(P ) and X(M ) are proved in the following way using the chain rule: T (ψ ◦ ϕ) ◦ X = T ψ ◦ T ϕ ◦ X = T ψ ◦ Y ◦ ϕ = Z ◦ ψ ◦ ϕ. The same chain rule calculation also shows that (iv) holds.



In this development we can replace F(M ) by C r (M, F) with little change; that is, we can replace realvalued functions by F-valued functions. Flows and Push-Forward and Pull-Back. Flows of vector fields behave naturally under the operations of ϕ-relatedness, push-forward and pull-back. This is what one would expect and is one of the indications that one is “doing things right”. 4.2.4 Proposition. Let ϕ : M → N be a C r -mapping of manifolds, X ∈ Xr (M ) and Y ∈ Xr (N ). Let FtX and FtY denote the flows of X and Y respectively. Then X ∼ϕ Y iff ϕ ◦ FtX = FtY ◦ ϕ. In particular, if ϕ is a diffeomorphism, then the equality Y = ϕ∗ X holds iff the flow of Y is ϕ ◦ FtX ◦ ϕ−1 . In particular, (FsX )∗ X = X. Proof. Taking the time derivative of the relation (ϕ ◦ FtX )(m) = (FtY ◦ ϕ)(m), for m ∈ M , using the chain rule and definition of the flow, we get  X  ∂FyY ∂Ft (m) Tϕ = (ϕ(m)), ∂t ∂t that is, (T ϕ ◦ X ◦ FtX )(m) = (Y ◦ FtY ◦ ϕ)(m) = (Y ◦ ϕ ◦ FtX )(m),

4.2 Vector Fields as Differential Operators

237

which is equivalent to T ϕ ◦ X = Y ◦ ϕ. Conversely, if this relation is satisfied, let c(t) = FtX (m) denote the integral curve of X through m ∈ M . Then   dc(t) d(ϕ ◦ c)(t) = Tϕ = T ϕ(X(c(t))) = Y ((ϕ ◦ c)(t)) dt dt says that ϕ ◦ c is the integral curve of Y through ϕ(c(0)) = ϕ(m). By uniqueness of integral curves, we get (ϕ ◦ FtX )(m) = (ϕ ◦ c)(t) = FtY (ϕ(m)). The last statement is obtained by taking ϕ = FsX for fixed s.  Note that for autonomous vector fields X the result of this proposition, namely (FsX )∗ X = X

(4.2.1)

may be directly proved by differentiating the flow identity Ft+s = Ft ◦ Fs with respect to s at s = 0. We call ϕ◦Ft ◦ϕ−1 the push-forward of Ft by ϕ since it is the natural way to construct a diffeomorphism on N out of one on M . See Figure 4.2.2. Thus, Proposition 4.2.4 says that the flow of the push-forward of a vector field is the push-forward of its flow .

X M

N

Figure 4.2.2. Pushing forward vector fields and integral curves.

Vector Fields as Operators on Functions. Next we define how vector fields operate on functions. This is done by means of the directional derivative. Let f : M → R, so T f : T M → T R = R × R. Recall that a tangent vector to R at a base point λ ∈ R is a pair (λ, µ), the number µ being the principal part. Thus we can write T f acting on a vector v ∈ Tm M in the form T f · v = (f (m), df (m) · v). ∗ This defines, for each m ∈ M , the element df (m) ∈ Tm M . Thus df is a section of T ∗ M , a covector field , or one-form.

4.2.5 Definition.

The covector field df : M → T ∗ M defined this way is called the differential of f .

For F-valued functions, f : M → F, where F is a Banach space, a similar definition gives df (m) ∈ L(Tm M, F) and we speak of df as an F-valued one-form. Clearly if f is C r , then df is C r−1 . Let us now work out df in local charts for f ∈ F(M ). If ϕ : U ⊂ M → V ⊂ E is a local chart for M , then the local representative of f is the map f : V → R defined by f = f ◦ ϕ−1 . The local representative of T f is the tangent map for local manifolds: T f (x, v) = (f (x), Df (x) · v).

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Thus, the local representative of df is the derivative of the local representative of f . In particular, if M is finite dimensional and local coordinates are denoted (x1 , . . . , xn ), then the local components of df are (df )i =

∂f ∂xi

The introduction of df leads to the following. 4.2.6 Definition. of f along X by

Let f ∈ F r (M ) and X ∈ Xr−1 (M ), r ≥ 1. Define the directional or Lie derivative £X f (m) ≡ X[f ](m) = df (m) · X(m),

for any m ∈ M . Denote by X[f ] = df (X) the map m ∈ M 7→ X[f ](m) ∈ R. If f is F-valued, the same definition is used, but now X[f ] is F-valued. The local representative of X[f ] in a chart is given by the function £X f (x) = Df (x) · X(x),

(4.2.2)

where f and X are the local representatives of f and X. In particular, if M is finite dimensional then we have n X ∂f i X . (4.2.3) X[f ] ≡ £X f = i ∂x i=1 Evidently if f is C r and X is C r−1 then X[f ] is C r−1 . From the chain rule, d(f ◦ ϕ) = df ◦ T ϕ, where ϕ : N → M is a C r map of manifolds, r ≥ 1. For real–valued functions, Leibniz’ rule gives d(f g) = f dg + gdf. (If f is F-valued, g is G-valued and B : F × G → H is a continuous bilinear map of Banach spaces, this generalizes to d(B(f, g)) = B(df, g) + B(f, dg).) Push-Forward and Lie Derivatives. It is natural to ask how Lie derivatives transform under mappings. No surprise: they transform in a natural way. 4.2.7 Proposition. (i) Suppose ϕ : M → N is a diffeomorphism. Then £X is natural with respect to push-forward by ϕ. That is, for each f ∈ F(M ), £ϕ∗ X (ϕ∗ f ) = ϕ∗ £X f ; in other words, the following diagram commutes:

F(M )

ϕ∗

£X ? F(M )

- F(N ) £ϕ∗ X

ϕ∗

? - F(N )

(ii) £X is natural with respect to restrictions. That is, for U open in M and f ∈ F(M ), £X|U (f |U ) = (£X f )|U ; or if |U : F(M ) → F(U ) denotes restriction to U , the following diagram commutes:

4.2 Vector Fields as Differential Operators

F(M )

|U

- F(U )

£X ? F(M )

Proof.

239

£X|U ? - F(U )

|U

For (i), if n ∈ N then £ϕ∗ X (ϕ∗ f )(n) = d(f ◦ ϕ−1 ) · (ϕ∗ X)(n) = d(f ◦ ϕ−1 )(n) · (T ϕ ◦ X ◦ ϕ−1 )(n) = df (ϕ−1 (n)) · (X ◦ ϕ−1 )(n) = ϕ∗ (£X f )(n).

(ii) follows from d(f |U ) = (df )|U , which itself is clear from the definition of d.



This proposition is readily generalized to F-valued C r functions. Since ϕ∗ = (ϕ−1 )∗ , the Lie derivative is also natural with respect to pull-back by ϕ. This has a generalization to ϕ-related vector fields as follows. 4.2.8 Proposition.

Let ϕ : M → N be a C r map, X ∈ Xr−1 (M ) and Y ∈ Xr−1 (N ). If X ∼ϕ Y , then £X (ϕ∗ f ) = ϕ∗ £Y f

for all f ∈ C r (N, F); that is, the following diagram commutes: C r (N, F)

ϕ∗

£Y ? C r−1 (N, F)

Proof.

ϕ∗

- C r (M, F) £X ? - C r−1 (M, F)

For m ∈ M , £X (ϕ∗ f )(m) = d(f ◦ ϕ)(m) · X(m) = df (ϕ(m)) · (Tm ϕ(X(m))) = df (ϕ(m)) · Y (ϕ(m)) = df (Y )(ϕ(m)) = (ϕ∗ £Y f )(m).

Properties of the Lie Derivative.



Next we show that £X satisfies the Leibniz rule.

4.2.9 Proposition. (i) The mapping £X : C r (M, F) → C r−1 (M, F) is a derivation. That is £X is R-linear and for f ∈ C r (M, F), g ∈ C r (M, G) and B : F × G → H a bilinear map £X (B(f, g)) = B(£X f, g) + B(f, £X g). In particular, for real-valued functions, £X (f g) = g£X f + f £X g. (ii) If c is a constant function, £X c = 0. Proof. Part (i) follows from the Leibniz rule for d and the definition £X f . Part (ii) results from the definition. 

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Lie Derivative Formula.

The connection between £X f and the flow of X is as follows.

4.2.10 Theorem (Lie Derivative Formula for Functions). Suppose f ∈ C r (M, F), X ∈ Xr−1 (M ), and X has a flow Ft . Then d ∗ F f = Ft∗ £X f dt t Proof. By the chain rule, the definition of the differential of a function and the flow of a vector field, together with equation (4.2.1), we have d ∗ d dFt (m) (Ft f )(m) = (f ◦ Ft )(m) = df (Ft (m)) · dt dt dt = df (Ft (m)) · X(Ft (m)) = df (X)(Ft (m)) = (£X f )(Ft (m)) = (Ft∗ £X f )(m).



As an application of the Lie derivative formula, we consider the problem of solving a partial differential equation on Rn+1 of the form n X ∂f ∂f (x, t) = X i (x) i (x, t) (4.2.4) ∂t ∂x i=1 with initial condition f (x, 0) = g(x) for given smooth functions X i (x), i = 1, . . . , n, g(x) and a scalar unknown f (x, t). 4.2.11 Proposition. Suppose X = (X 1 , . . . , X n ) has a complete flow Ft . Then f (x, t) = g(Ft (x)) is a solution of the foregoing problem (4.2.4). (See Exercise 4.2-4 for uniqueness.) Proof. give

To prove the theorem we take the time derivative of f (x, t) and use the Lie derivative formula to ∂f d = Ft∗ g = Ft∗ £X g = £X (Ft∗ g) = X[f ]. ∂t dt

which proves the theorem.



Thus, one can solve this scalar equation by computing the orbits of X and pushing (or “dragging along”) the graph of g by the flow of X; see Figure 4.2.3. These trajectories of X are called characteristics of (P). (As we shall see below, the vector field X in (P) can be time dependent.)

Figure 4.2.3. Solving a PDE using characteristics.

4.2.12 Example.

Solve the partial differential equation   ∂f ∂f ∂f = (x + y) − , ∂t ∂x ∂y

with initial condition f (x, y, 0) = x2 + y 2 .



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241

Solution. The vector field X(x, y) = (x + y, −x − y) has a complete flow Ft (x, y) = ((x + y)t + x, −(x + y)t + y), so that the solution of the previous partial differential equation is given by f (x, y, t) = 2(x + y)2 t2 + x2 + y 2 + 2(x2 − y 2 )t.



Differential Operators are Vector Fields. Now we turn to the question of using the Lie derivative to characterize vector fields. We will prove that any derivation on functions uniquely defines a vector field. Because of this, derivations can be (and often are) used to define vector fields. (See the introduction to §3.3.) In the proof we shall need to localize things in a smooth way, hence the following lemma of general utility is proved first. 4.2.13 Lemma. Let E be a C r Banach space, that is, one whose norm is C r on E\{0}, r ≥ 1. Let U1 be an open ball of radius r1 about x0 and U2 an open ball of radius r2 , r1 < r2 . Then there is a C r function h : E → R such that h is one on U1 and zero outside U2 . We call h a bump function. Later we will prove more generally that on a manifold M , if U1 and U2 are two open sets with cl(U1 ) ⊂ U2 , there is an h ∈ F r (M ) such that h is one on U1 and is zero outside U2 . Proof. By a scaling and translation, we can assume that U1 and U2 are balls of radii 1 and 3 and centered at the origin. Let θ : R → R be given by   −1 if |x| < 1 θ(x) = exp 1 − |x|2 and set if |x| ≥ 1.

θ(x) = 0, (See the remarks following Theorem 2.4.15.) Now set Z

s

θ(t) dt θ1 (s) =

Z−∞ ∞ θ(t) dt −∞

so θ1 is a C ∞ function and is 0 if s < −1, and 1 if s > 1. Let θ2 (s) = θ1 (2 − s), so θ2 is a C ∞ function that is 1 if s < 1 and 0 if s > 3. Finally, let h(x) = θ2 (kxk).  The norm on a real Hilbert space is C ∞ away from the origin. The order of differentiability of the norms of some concrete Banach spaces is also known; see Bonic and Frampton [1966], Yamamuro [1974], and Supplement 5.5B. ∗ 4.2.14 Corollary. Let M be a C r manifold modeled on a C r Banach space. If αm ∈ Tm M , then there is r an f ∈ F (M ) such that df (m) = αm .

Proof. If M = E, so Tm E ∼ = E, let f (x) = αm (x), a linear function on E. Then df is constant and equals αm . The general case can be reduced to E using a local chart and a bump function as follows. Let ϕ : U → U 0 ⊂ E be a local chart at m with ϕ(m) = 0 and such that U 0 contains the ball of radius 3. Let α ˜ m be the local representative of αm and let h be a bump function, 1 on the ball of radius 1 and zero outside the ball of radius 2. let f (x) = α ˜ m (x) and let ( (hf ) ◦ ϕ, on U f= 0, on M \U. It is easily verified that f is C r and df (m) = αm .



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4. Vector Fields and Dynamical Systems

4.2.15 Proposition. (i) Let M be a C r manifold modeled on a C r Banach space. The collection of operators £X for X ∈ Xr (M ), defined on C r (M, F ) and taking values in C r−1 (M, F ) forms a real vector space and an F r (M )-module with (f £X )(g) = f (£X g), and is isomorphic to Xr (M ) as a real vector space and as an F r (M )-module. In particular, £X = 0 iff X = 0; and £f X = f £X . (ii) Let M be any C r manifold. If £X f = 0 for all f ∈ C r (U, F ), for all open subsets U of M , then X = 0. Proof.

(i) Consider the map σ : X 7→ £X . It is obviously R and F r (M ) linear; that is, £X1 +f X2 = £X1 + f £X2 .

To show that it is one–to–one, we must show that £X = 0 implies X = 0. But if £X f (m) = 0, then ∗ df (m) · X(m) = 0 for all f . Hence, αm (X(m)) = 0 for all αm ∈ Tm M by Corollary 4.2.14. Thus X(m) = 0 by the Hahn–Banach theorem. (ii) This has an identical proof with the only exception that one works in a local chart, so it is not necessary to extend a linear functional to the entire manifold M as in Corollary 4.2.14. Thus the condition on the differentiability of the norm of the model space of M can be dropped.  4.2.16 Theorem (Derivation Theorem). (i) If M is finite dimensional and C ∞ , the collection of all derivations on F(M ) is a real vector space isomorphic to X(M ). In particular, for each derivation θ there is a unique X ∈ X(M ) such that θ = £X . (ii) Let M be a C ∞ manifold modeled on a C ∞ Banach space E, that is, E has a C ∞ norm away from the origin. The collection of all (R-linear) derivations on C ∞ (M, F) (for all Banach spaces F) forms a real vector space isomorphic to X(M ). Proof. We prove (ii) first. Let θ be a derivation. We wish to construct X such that θ = £X . First of all, note that θ is a local operator ; that is, if h ∈ C ∞ (M, F) vanishes on a neighborhood V of m, then θ(h)(m) = 0. Indeed, let g be a bump function equal to one on a neighborhood of m and zero outside V . Thus h = (1 − g)h and so θ(h)(m) = θ(1 − g)(m) · h(m) + θ(h)(m)(1 − g(m)) = 0.

(4.2.5)

If U is an open set in M , and f ∈ C ∞ (U, F) define (θ|U )(f )(m) = θ(gf )(m), where g is a bump function equal to one on a neighborhood of m and zero outside U . By the previous remark, (θ|U )(f )(m) is independent of g, so θ|U is well defined. For convenience we write θ = θ|U . Let (U, ϕ) be a chart on M , m ∈ U , and f ∈ C ∞ (M, F) where ϕ : U → U 0 ⊂ E; we can write, for x ∈ U 0 and a = ϕ(m), Z 1 ∂ (ϕ∗ f )[a + t(x − a)] dt (ϕ∗ f )(x) = (ϕ∗ f )(a) + ∂t 0 Z 1 = (ϕ∗ f )(a) + D(ϕ∗ f )[a + t(x − a)] · (x − a) dt 0

This formula holds in some neighborhood ϕ(V ) of a. Hence for u ∈ V we have f (u) = f (m) + g(u) · (ϕ(u) − a), where g ∈ C ∞ (V, L(E, F)) is given by Z

1

D(ϕ∗ f )[a + t(ϕ(u) − a)]dt.

g(u) = 0

(4.2.6)

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243

Applying θ to equation (4.2.6) at u = m gives θf (m) = g(m) · (θϕ)(m) = D(ϕ∗ f )(a) · (θϕ)(m).

(4.2.7)

Since θ was given globally, the right hand side of equation (4.2.7) is independent of the chart. Now define X on U by its local representative Xϕ (x) = (x, θ(ϕ)(u)), where x = ϕ(u) ∈ U 0 . It follows that X|U is independent of the chart ϕ and hence X ∈ X(M ). Then, for f ∈ C ∞ (M, F), the local representative of £X f is D(f ◦ ϕ−1 )(x) · Xϕ (x) = D(f ◦ ϕ−1 )(x) · (θϕ)(u) = θf (u). Hence £X = θ. Finally, uniqueness follows from Proposition 4.2.15. The vector derivative property was used only in establishing equations (4.2.5) and (4.2.7). Thus, if M is finite dimensional and θ is a derivation on F(M ), we have as before f (u) = f (m) + g(u) · (ϕ(u) − a) = f (m) +

n X

(ϕi (u) − ai )gi (u),

i=1

where gi ∈ F(V ) and ∂(ϕ∗ f )(u) , gi (m) = ∂X i u=a

a = (a1 , . . . , an ).

Hence equation (4.2.7) becomes θf (m) =

n X i=1

gi (m) θ(ϕi )(m) =

n X ∂ (ϕ∗ f )(a) θ(ϕi )(m) i ∂x i=1

and this is again independent of the chart. Now define X on U by its local representative (x, θ(ϕ1 )(u), . . . , θ(ϕn )(u)) and proceed as before.



Remark. There is a difficulty with this proof for C r manifolds and derivations mapping C r to C r−1 . Indeed in equation (4.2.6), g is only C r−1 if f is C r , so θ need not be defined on g. Thus, one has to regard θ as defined on C r−1 -functions and taking values in C r−2 -functions. Therefore θ(ϕ) is only C r−1 and so the vector field it defines is also only C r−2 . Then the above proof shows that £X and θ coincide on C r -functions on M . In Supplement 4.2D we will prove that £X and θ are in fact equal on C r−1 -functions, but the proof requires a different argument.  For finite-dimensional manifolds, the preceding theorem provides a local basis for vector fields. If (U, ϕ), ϕ : U → V ⊂ Rn is a chart on M defining the coordinate functions xi : U → R, define n derivations ∂/∂xi on F(U ) by ∂f ∂(f ◦ ϕ−1 ) = ◦ ϕ. i ∂x ∂xi These derivations are linearly independent with coefficients in F(U ), for if ! n n X X i ∂ i ∂ f = 0, then f (xj ) = f j = 0 i i ∂x ∂x i=1 i=1

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for all j = 1, . . . , n, since (∂/∂xi )xj = δij . By Theorem 4.2.16, (∂/∂xi ) can be identified with vector fields on U . Moreover, if X ∈ X(M ) has components X 1 , . . . , X n in the chart ϕ, then ! n n n X X X ∂ i ∂ i ∂f = X f, i.e., X = Xi i £X f = X[f ] = X i i ∂x ∂x ∂x . i=1 i=1 i=1 Thus the vector fields (∂/∂xi ), i = 1, . . . , n form a local basis for the vector fields on M . It should be mentioned however that a global basis of X(M ), that is, n vector fields, X1 , . . . , Xn ∈ X(M ) that are linearly independent over F(M ) and span X(M ), does not exist in general. Manifolds that do admit such a global basis for X(M ) are called parallelizable. It is straightforward to show that a finite-dimensional manifold is parallelizable iff its tangent bundle is trivial. For example, it is shown in differential topology that S 3 is parallelizable but S 2 is not (see Supplement 7.5A). This completes the discussion of the Lie derivative on functions. Turning to the Lie derivative on vector fields, let us begin with the following. 4.2.17 Proposition.

If X and Y are C r vector fields on M , then [£X , £Y ] = £X ◦ £Y − £Y ◦ £X

is a derivation mapping C r+1 (M, F) to C r−1 (M, F). Proof. More generally, let θ1 and θ2 be two derivations mapping C r+1 to C r and C r to C r−1 . Clearly [θ1 , θ2 ] = θ1 ◦ θ2 − θ2 ◦ θ1 is linear and maps C r+1 to C r−1 . Also, if f ∈ C r+1 (M, F), g ∈ C r+1 (M, G), and B ∈ L(F, G; H), then [θ1 , θ2 ] (B(f, g)) = (θ1 ◦ θ2 )(B(f, g)) − (θ2 ◦ θ1 )(B(f, g)) = θ1 {B(θ2 (f ), g) + B(f, θ2 (g))} − θ2 {B(θ1 (f ), g) + B(f, θ1 (g))} = B(θ1 (θ2 (f )), g) + B(θ2 (f ), θ1 (g)) + B(θ1 (f ), θ2 (g)) + B(f, θ1 (θ2 (g))) − B(θ2 (θ1 (f )), g) − B(θ1 (f ), θ2 (g)) − B(θ2 (f ), θ1 (g)) − B(f, θ2 (θ1 (g))) = B([θ1 , θ2 ] (f ), g) + B(f, [θ1 , θ2 ] (g)).  The Jacobi-Lie Bracket of Two Vector Fields. can be given.

Because of Theorem 4.2.16 the following definition

4.2.18 Definition. Let M be a manifold modeled on a C ∞ Banach space and X, Y ∈ X∞ (M ). Then [X, Y ] is the unique vector field such that £[X,Y ] = [£X , £Y ]. This vector field is also denoted £X Y and is called the Lie derivative of Y with respect to X, or the Jacobi–Lie bracket of X and Y . Even though this definition is useful for Hilbert manifolds (in particular for finite-dimensional manifolds), it excludes consideration of C r vector fields on Banach manifolds modeled on nonsmooth Banach spaces, such as Lp function spaces for p not even. We shall, however, establish an equivalent definition, which makes sense on any Banach manifold and works for C r vector fields. This alternative definition is based on the following result. 4.2.19 Theorem (Lie Derivative Formula for Vector Fields). X(M ), and let X have (local ) flow Ft . Then d ∗ (F Y ) = Ft∗ (£X Y ) dt t (at those points where Ft is defined ).

Let M be as in Definition 4.2.18, X, Y ∈

4.2 Vector Fields as Differential Operators

245

Note that as a consequence of this result and using Proposition 4.2.7 together with the identity Ft∗ X = X (equation (4.2.1)), we have Ft∗ (£X Y ) = £X Ft∗ Y Because the equation Ft∗ X = X is only valid for autonomous vector fields, we shall have to revisit these questions for the case of time varying vector fields. Proof.

If t = 0 this formula becomes d F ∗ Y = £X Y. dt t=0 t

(4.2.8)

Assuming equation (4.2.8) for the moment, d ∗ d ∗ ∗ d F Y = Ft F ∗ Y = Ft∗ £X Y. (F Y ) = dt t ds s=0 t+s ds s=0 s Thus the formula in the theorem is equivalent to equation (4.2.8), which is proved in the following way. Both sides of equation (4.2.8) are clearly vector derivations. In view of Theorem 4.2.16, it suffices then to prove that both sides are equal when acting on an arbitrary function f ∈ C ∞ (M, F). Now d d ∗ ∗ f )]}(m) (F Y )[f ](m) = {£Ft∗ Y [Ft∗ (F−t dt t=0 t dt t=0 d ∗ = F ∗ (Y [F−t f ])(m), dt t=0 t by Proposition 4.2.7(i). Using Theorem 4.2.10 and Leibniz’ rule, this becomes X[Y [f ]](m) − Y [X[f ]](m) = [X, Y ][f ](m), which proves the theorem.



Since the formula for £X Y in equation (4.2.8) does not use the fact that the norm of E is C ∞ away from the origin, we can state the following definition of the Lie derivative on any Banach manifold M . 4.2.20 Definition (Dynamic Definition of Jacobi–Lie bracket). If X, Y ∈ Xr (M ), r ≥ 1 and X has flow Ft , the C r−1 vector field £X Y = [X, Y ] on M defined by d [X, Y ] = (F ∗ Y ) dt t=0 t is called the Lie derivative of Y with respect to X, or the Lie bracket of X and Y . Theorem 4.2.19 then shows that this definition agrees with the earlier one, Definition 4.2.18: 4.2.21 Proposition. M satisfying

Let X, Y ∈ Xr (M ), r ≥ 1. Then [X, Y ] = £X Y is the unique C r−1 vector field on [X, Y ][f ] = X[Y [f ]] − Y [X[f ]]

for all f ∈ C r+1 (U, F), where U is open in M . The derivation approach suggests that if X, Y ∈ Xr (M ) then [X, Y ] might only be C r−2 , since [X, Y ] maps C r+1 functions to C r−1 functions, and differentiates them twice. However Definition 4.2.20 (and the local expression equation (4.2.10) below) show that [X, Y ] is in fact C r−1 .

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The Vector Fields as a Lie Algebra. fields a very interesting structure.

The Jacobi-Lie bracket operation gives the space of all vector

4.2.22 Proposition. The bracket [X, Y ] on X(M ), together with the real vector space structure X(M ), form a Lie algebra. That is, (i) [ , ] is R bilinear ; (ii) [X, X] = 0 for all X ∈ X(M ); (iii) [X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y ]] = 0 for all X, Y, Z ∈ X(M ) (Jacobi identity ). The proof is straightforward, applying the brackets in question to an arbitrary function. Unlike X(M ), the space Xr (M ) is not a Lie algebra since [X, Y ] ∈ Xr−1 (M ) for X, Y ∈ Xr (M ). (i) and (ii) imply that [X, Y ] = −[Y, X], since [X + Y, X + Y ] = 0 = [X, X] + [X, Y ] + [Y, X] + [Y, Y ]. We can describe (iii) by writing £X as a Lie bracket derivation: £X [Y, Z] = [£X Y, Z] + [Y, £X Z]. Strictly speaking we should be careful using the same symbol £X for both definitions of £X f and £X Y . However, the meaning is generally clear from the context. Push-Forward of Brackets.

The analog of Proposition 4.2.7 on the vector field level is the following.

4.2.23 Proposition. (i) Let ϕ : M → N be a diffeomorphism and X ∈ X(M ). Then £X : X(M ) → X(M ) is natural with respect to push-forward by ϕ. That is, £ϕ∗ X ϕ∗ Y = ϕ∗ £X Y, i.e., [ϕ∗ X, ϕ∗ Y ] = ϕ∗ [X, Y ], i.e., the following diagram commutes: X(M )

ϕ∗

£X

- X(N ) £ϕ∗ X ? - X(N )

? X(M ) ϕ∗

(ii) £X is natural with respect to restrictions. That is, for U ⊂ M open, [X|U, Y |U ] = [X, Y ]|U ; or the following diagram commutes: X(M )

|U

£X

- X(U ) £X|U ? - X(U )

? X(M ) |U

Proof. For (i), let f ∈ F(V ), V be open in N , and ϕ(m) = n ∈ V . By Proposition 4.2.7(i), for any Z ∈ X(M ) ((ϕ∗ Z)[f ])(n) = Z[f ◦ ϕ](m),

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247

so we get from Proposition 4.2.21 (ϕ∗ [X, Y ])[f ](n) = [X, Y ][f ◦ ϕ](m) = X[Y [f ◦ ϕ]](m) − Y [X[f ◦ ϕ]](m) = X[(ϕ∗ Y )[f ] ◦ ϕ](m) − Y [(ϕ∗ X)[f ] ◦ ϕ](m) = (ϕ∗ X)[(ϕ∗ Y )[f ]](n) − (ϕ∗ Y )[(ϕ∗ X)[f ]](n) = [ϕ∗ X, ϕ∗ Y ][f ](n). Thus, ϕ∗ [X, Y ] = [ϕ∗ X, ϕ∗ Y ] by Proposition 4.2.15(ii). (ii) follows from the fact that d(f |U ) = df |U .



The Coordinate Expression for the Jacobi-Lie Bracket. Let us compute the local expression for [X, Y ]. Let ϕ : U → V ⊂ E be a chart on M and let the local representatives of X and Y be denoted X and Y respectively, so X, Y : V → E. By Proposition 4.2.23, the local representative of [X, Y ] is given by [X, Y]. Thus, [X, Y][f ](x) = X[Y[f ]](x) − Y[X[f ]](x) = D(Y[f ])(x) · X(x) − D(X[f ])(x) · Y(x). Now Y[f ](x) = Df (x) · Y(x) and its derivative may be computed by the product rule. The terms involving the second derivative of f cancel by symmetry of D2 f (x) and so we are left with Df (x) · {DY(x) · X(x) − DX(x) · Y(x)}. Thus the local representative of [X, Y] is [X, Y] = DY · X − DX · Y.

(4.2.9)

If M is n-dimensional and the chart ϕ gives local coordinates (x1 , . . . , xn ) then this calculation gives the components of [X, Y ] as n X ∂X j ∂Y j j (4.2.10) [X, Y ] = Xi i − Y i ∂x ∂xi i=1 that is, [X, Y] = (X · ∇)Y − (Y · ∇)X. The formula (4.2.10) can be easily remembered by the following device: Let X = X i ∂/∂xi and Y = j Y ∂/∂xj and then write   ∂ ∂ [X, Y] = X i i , Y j j ∂x ∂x and treat the expression on the right hand side as a commutator in the usual way, but ignoring the second derivative terms (which will cancel anyway, due to equality of mixed partial derivatives). ϕ-Relatedness of Brackets. Part (i) of Proposition 4.2.23 has an important generalization to ϕ-related vector fields. For this, however, we need first the following preparatory proposition. 4.2.24 Proposition. Let ϕ : M → N be a C r map of C r manifolds, X ∈ Xr−1 (M ), and X 0 ∈ Xr−1 (N ). Then X ∼ϕ X 0 iff (X 0 [f ]) ◦ ϕ = X[f ◦ ϕ] for all f ∈ F 1 (V ), where V is open in N . Proof.

By definition, ((X 0 [f ]) ◦ ϕ)(m) = df (ϕ(m)) · X 0 (ϕ(m)). By the chain rule, X[f ◦ ϕ](m) = d(f ◦ ϕ)(m) · X(m) = df (ϕ(m)) · Tm ϕ(X(m)).

If X ∼ϕ X 0 , then T ϕ ◦ X = X 0 ◦ ϕ and we have the desired equality. Conversely, if X[f ◦ ϕ] = (X 0 [f ]) ◦ ϕ for all f ∈ F 1 (V ), and all V open in N , choosing V to be a chart domain and f the pull-back to V of linear functionals on the model space of N , we conclude that αn · (X 0 ◦ ϕ)(m) = αn · (T ϕ ◦ X)(m), where n = ϕ(m), for all αn ∈ Tn∗ N . Using the Hahn–Banach theorem, we deduce that (X 0 ◦ ϕ)(m) = (T ϕ ◦ X)(m), for all m ∈ M. 

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It is to be noted that under differentiability assumptions on the norm on the model space of N (as in Theorem 4.2.16), the condition “for all f ∈ F 1 (V ) and all V ⊂ N ” can be replaced by “for all f ∈ F 1 (N )” by using bump functions. This holds in particular for Hilbert (and hence for finite-dimensional) manifolds. 4.2.25 Proposition. Let ϕ : M → N be a C r map of manifolds, X, Y ∈ Xr−1 (M ), and X 0 , Y 0 ∈ Xr−1 (N ). If X ∼ϕ X 0 and Y ∼ϕ Y 0 , then [X, Y ] ∼ϕ [X 0 , Y 0 ]. Proof. By Proposition 4.2.24 it suffices to show that ([X 0 , Y 0 ][f ]) ◦ ϕ = [X, Y ][f ◦ ϕ] for all f ∈ F 1 (V ), where V is open in N . To show this, note that ([X 0 , Y 0 ][f ]) ◦ ϕ = X 0 [Y 0 [f ]] ◦ ϕ − Y 0 [X 0 [f ]] ◦ ϕ = X[(Y 0 [f ]) ◦ ϕ] − Y [(X 0 [f ]) ◦ ϕ] = X[Y [f ◦ ϕ]] − Y [X[f ◦ ϕ]] = [X, Y ][f ◦ ϕ], which proves the result.

(4.2.11) 

Further Properties of the Lie Derivative.

The analog of Proposition 4.2.9 is the following.

4.2.26 Proposition. For every X ∈ X(M ), the operator £X is a derivation on (F(M ), X(M )). That is, £X is R-linear and £X (f Y ) = (£X f )Y + f (£X Y ). Proof.

For g ∈ C ∞ (U, E), where U is open in M , we have [X, f Y ][g] = £X (£f Y g) − £f Y £X g = £X (f £Y g) − f £Y £X g = (£X f )£Y g + f £X £Y g − f £Y £X g,

so [X, f Y ] = (£X f ) Y + f [X, Y ], by Proposition 4.2.15(ii).

(4.2.12) 

Commutation of vector fields is characterized by their flows in the following way. 4.2.27 Proposition. equivalent.

Let X, Y ∈ Xr (M ), r ≥ 1, and let Ft , Gt denote their flows. The following are

(i) [X, Y ] = 0; (ii) Ft∗ Y = Y ; (iii) G∗t X = X; (iv) Ft ◦ Gs = Gs ◦ Ft . (In (ii)–(iv), equality is understood, as usual, where the expressions are defined.) Proof. Ft ◦ Gs = Gs ◦ Ft iff Gs = Ft ◦ Gs ◦ Ft−1 , which by Proposition 4.2.4 is equivalent to Y = Ft∗ Y ; that is, (iv) is equivalent to (ii). Similarly (iv) is equivalent to (iii). If Ft∗ Y = Y , then d F ∗ Y = 0. [X, Y ] = dt t=0 t Conversely, if [X, Y ] = £X Y = 0, then d d ∗ F Y = F ∗ Y = Ft∗ [X, Y ] = 0 dt t ds s=0 t+s so that Ft∗ Y is constant in t. For t = 0, however, its value is Y , so that Ft∗ Y = Y and we have thus showed that (i) and (ii) are equivalent. Similarly (i) and (iii) are equivalent. 

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Solving Systems of PDE’s. Just as in Theorem 4.2.10, the formula for the Lie derivative involving the flow can be used to solve special types of first-order linear n × n systems of partial differential equations. Consider the first-order system:  n  X ∂X i (x) ∂Y i ∂Y i (x, t) = X j (x) j (x, t) − Y j (x, t) ∂t ∂x ∂xj j=1

(4.2.13)

with initial conditions Y i (x, 0) = g i (x) for given functions X i (x), g i (x) and scalar unknowns Y i (x, t), i = 1, . . . , n, where x = (x1 , . . . , xn ). 4.2.28 Proposition. Suppose X = (X 1 , . . . , X n ) has a complete flow Ft . Then letting Y = (Y 1 , . . . , Y n ) and G = (g 1 , . . . , g n ), Y = Ft∗ G is a solution of the problem (4.2.13). (See Exercise 4.2-4 for uniqueness.) Proof.

d ∂Y = Ft∗ G = Ft∗ [X, G] = [Ft∗ X, Ft∗ G] = [X, Y ] ∂t dt since Ft∗ X = X and Y = Ft∗ G. The expression in the problem (Pn ) is by equation (4.2.10) the ith component of [X, Y ].  4.2.29 Example.

Solve the system of partial differential equations: ∂Y 1 ∂Y 1 ∂Y 1 = (x + y) − (x + y) − Y 1 − Y 2, ∂t ∂x ∂y ∂Y 2 ∂Y 2 ∂Y 2 = (x + y) − (x + y) +Y1+Y2 ∂t ∂x ∂y

with initial conditions Y 1 (x, y, 0) = x, Y 2 (x, y, 0) = y 2 . The vector field X(x, y) = (x + y, −x − y) has the complete flow Ft (x, y) = ((x + y)t + x, −(x + y)t + y), so that the solution is given by Y (x, y, t) = Ft∗ (x, y 2 ); that is, Y 1 (x, y, t) = ((x + y)t + x)(1 − t) − t[y − (x + y)t]2 Y 2 (x, y, t) = t((x + y)t + x) + (t + 1)[y − (x + y)t]2 .



Time Dependent Vector Fields. In later chapters we will need a flow type formula for the Lie derivative of a time-dependent vector field. Recall that in §4.1 we discussed the existence and uniqueness of solutions of a time-dependent vector field. Let us formalize and recall the basic facts. 4.2.30 Definition. A C r time-dependent vector field is a C r map X : R × M → T M such that X(t, m) ∈ Tm M for all (t, m) ∈ R × M ; that is, Xt ∈ Xr (M ), where Xt (m) = X(t, m). The timedependent flow or evolution operator Ft,s of X is defined by the requirement that t 7→ Ft,s (m) be the integral curve of X starting at m at time t = s; that is, d Ft,s (m) = X(t, Ft,s (m)) dt

and

Fs,s (m) = m.

By uniqueness of integral curves we have Ft,s ◦ Fs,r = Ft,r (replacing the flow property Ft+s = Ft ◦ Fs ), and Ft,t = identity. It is customary to write Ft = Ft,0 . If X is time independent, Ft,s = Ft−s . In general Ft∗ Xt 6= Xt . However, the basic Lie derivative formulae still hold. 4.2.31 Theorem. Let Xt ∈ Xr (M ), r ≥ 1 for each t and suppose X(t, m) is continuous in (t, m). Then Ft,s is of class C r and for f ∈ C r+1 (M, F ), and Y ∈ Xr (M ), we have (i)

d ∗ ∗ F f = Ft,s (£Xt f ), and dt t,s

250

(ii)

4. Vector Fields and Dynamical Systems

d ∗ ∗ ∗ F Y = Ft,s ([Xt , Y ]) = Ft,s (£Xt Y ). dt t,s

Proof.

That Ft,s is C r was proved in §4.1. The proof of (i) is a repeat of Theorem 4.2.10: d ∗ d (F f )(m) = (f ◦ Ft,s )(m) dt t,s dt dFt,s (m) = df (Ft,s (m)) dt = df (Ft,s (m)) · Xt (Ft,s (m)) = (£Xt f )(Ft,s (m)) ∗ = Ft,s (£Xt f )(m).

For vector fields, note that by Proposition 4.2.7(i), ∗ ∗ ∗ (Ft,s Y )[f ] = Ft,s (Y [Ft,s f ])

(4.2.14)

−1 since Fs,t = Ft,s . The result (ii) will be shown to follow from (i), equation (4.2.14), and the next lemma.

4.2.32 Lemma.

The following identity holds: d ∗ ∗ F f = −Xt [Fs,t f ]. dt s,t

Proof.

Differentiating Fs,t ◦ Ft,s = identity in t, we get the backward differential equation: d Fs,t = −T Fs,t ◦ X dt

Thus d ∗ F f (m) = −df (Fs,t (m)) · T Fs,t (Xt (m)) dt s,t = −dF (f ◦ Fs,t ) · Xt (m) = −Xt [f ◦ Fs,t ](m).

H

Thus from equation (4.2.14) and (i), d ∗ ∗ ∗ ∗ ∗ (F Y )[f ] = Ft,s (Xt [Y [Ft,s f ]]) − Ft,s (Y [Xt [Ft,s f ]]), dt t,s ∗ By Proposition 4.2.21 and equation (4.2.14), this equals (Ft,s [Xt , Y ])[f ].

If f and Y are time dependent, then (i) and (ii) read   d ∗ ∂f ∗ F f = Ft,s + £Xt f dt t,s ∂t and d ∗ ∗ F Y = Ft,s dt t,s



 ∂Y + [Xt , Yt ] . ∂t

Unlike the corresponding formula for time-independent vector fields, we generally have ∗ ∗ Ft,s (£Xt f ) 6= £Xt (Ft,s f)

and

∗ ∗ ∗ X Xt (F Ft,s (£Xt Y ) 6= £Ft,s t,s Y ). t



(4.2.15)

(4.2.16)

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The Suspension Trick. Time-dependent vector fields on M can be made into time-independent ones on a bigger manifold. Let t ∈ X(R × M ) denote the vector field which is defined by t(s, m) = ((s, 1), 0m ) ∈ T(s,m) (R × M ) ∼ = Ts R × Tm M . Let the suspension of X be the vector field X 0 ∈ X(R × M ) where 0 X (t, m) = ((t, 1), X(t, m)) and observe that X 0 = t + X. Since b : I → M is an integral curve of X at m iff b0 (t) = X(t, b(t)) and b(0) = m, a curve c : I → R × M is an integral curve of X 0 at (0, m) iff c(t) = (t, b(t)). Indeed, if c(t) = (a(t), b(t)) then c(t) is an integral curve of X 0 iff c0 (t) = (a0 (t), b0 (t)) = X 0 (c(t)); that is a0 (t) = 1 and b0 (t) = X(a(t), b(t)). Since a(0) = 0, we get a(t) = t. These observations are summarized in the following (see Figure 4.2.4).

M m

integral curve of X

Figure 4.2.4. Suspension of a vector field.

4.2.33 Proposition. Let X be a C r -time-dependent vector field on M with evolution operator Ft,s . The flow Ft of the suspension X 0 ∈ Xr (R × M ) is given by Ft (s, m) = (t + s, Ft+s,s (m)). Proof. In the preceding notations, b(t) = Ft,0 (m), c(t) = Ft (0, m) = (t, Ft,0 (m)), and so the statement is proved for s = 0. In general, note that Ft (s, m) = Ft+s (0, F0,s (m)) since t 7→ Ft+s (0, F0,s (m)) is the integral curve of X 0 , which at t = 0 passes through Fs (0, F0,s (m)) = (s, (Fs,0 ◦ F0,s )(m)) = (s, m). Thus, Ft (s, m) = Ft+s (0, F0,s (m)) = (t + s, (Ft+s,0 ◦ F0,s )(m)) = (t + s, Ft+s,s (m)), as required.



Supplement 4.2A Product formulas for the Lie bracket This supplement is a continuation of Supplement 4.1A and gives the flow of the Lie bracket [X, Y ] in terms of the flows of the vector fields X, Y ∈ X(M ). 4.2.34 Proposition. Let X, Y ∈ X(M ) have flows Ft and Gt . If Bt denotes the flow of [X, Y ], then for x ∈ M,  n Bt (x) = lim G−√t/n ◦ F−√t/n ◦ G√t/n ◦ F√t/n (x), t ≥ 0. n→∞

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Proof.

Let
Kε (x) = G−√ε ◦ F−√ε ◦ G√ε ◦ F√ε (x),

ε ≥ 0.

The claimed formula follows from Theorem 4.1.26 if we show that ∂ = [X, Y ](x) Kε (x) ∂ε ε=0 for all x ∈ M . This in turn is equivalent to ∂ ∗ = [X, Y ](x) K f ∂ε ε ε=0 for any f ∈ F(M ). By the Lie derivative formula,     ∂ ∗ 1 n ∗ ∗ ∗ ∗ √ F ∗ √ G∗ √ f + F√ √ £Y F ∗ √ G∗ √ f Kε f = √ F√ £ G G X − ε − ε ε ε − ε − ε ε ε ∂ε 2 ε  o  ∗ ∗ ∗ ∗ √ F ∗ √ £X G∗ √ f − F√ √ F ∗ √ G∗ √ £Y (f ) . −F√ G G ε ε − ε ε ε − ε − ε − ε √ By the chain rule, the limit of this as ε ↓ 0 is half the ∂/∂ ε-derivative of the parenthesis at ε = 0. Again by the Lie derivative formula, this equals a sum of 16 terms, which reduces to the expression [X, Y ][f ].  For example, for n × n matrices A and B, Proposition 4.2.34 yields the classical formula  √ √ √ √ n e[A,B] = lim e−A/ n e−B/ n eA/ n eB/ n , n→∞

where the commutator is given by [A, B] = AB − BA. To see this, define for any n × n matrix C a vector field XC ∈ X(Rn ) by XC (x) = Cx. Thus XC has flow Ft (x) = etC x. Note that XC is linear in C and satisfies [XA , XB ] = −X[A,B] as is easily verified. Thus the flow of [XB , XA ] is et[A,B] and the formula now follows from Proposition 4.2.34. The results of Corollary 4.1.27 and Proposition 4.2.34 had their historical origins in Lie group theory, where they are known by the name of exponential formulas. The converse of Corollary 4.1.25, namely expressing etA etB as an exponential of some matrix for sufficiently small t is the famous Baker–Campbell–Hausdorff formula (see e.g., Varadarajan [1974, Section 2.15]). The formulas in Corollary 4.1.25 and Proposition 4.2.34 have certain generalizations to unbounded operators and are called Trotter product formulas after Trotter [1958]. See Chorin, Hughes, McCracken, and Marsden [1978] for further information.

Supplement 4.2B The Method of Characteristics The method used to solve problem (4.2.4) also enables one to solve first-order quasi-linear partial differential equations in Rn . Unlike Proposition 4.2.11, the solution will be implicit, not explicit. The equation under consideration on Rn is n X ∂f X i (x1 , . . . , xn , f ) i = Y (x1 , . . . , xn , f ), (4.2.17) ∂x i=1

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where f = f (x1 , . . . , xn ) is the unknown function and X i , Y , i = 1, . . . , n are C r real-valued functions on Rn+1 , r ≥ 1. As initial condition one takes an (n − 1)-dimensional submanifold Γ in Rn+1 that is nowhere tangent to the vector field n X ∂ ∂ Z := Xi i + Y ∂x ∂f i=1 called the characteristic vector field of (4.2.17). We assume that Γ is given parametrically in terms of an injective immersion as follows. The manifold Γ is the image of a map of Rn−1 to Rn+1 that we write as xi = ξ0i (t1 , . . . , tn−1 ), i = 1, . . . , n

and

f = ϕ0 (t1 , . . . , tn−1 ).

We require that the initial conditions hold on Γ in the sense that f at the points of Γ be given by ϕ0 ; namely,
f ξ01 (t1 , . . . , tn−1 ), . . . ξ0n (t1 , . . . , tn−1 ) = ϕ0 (t1 , . . . , tn−1 ) for all (t1 , . . . , tn−1 ) in the domain of the immersion.  X1 ···   ∂ξ 1  0 ···   ∂t1  .  ..   ∂ξ 1 0 ··· ∂tn−1

The non-tangency requirement means that the matrix  Xn Y  ∂ξ0n ∂ϕ0    ∂t1 ∂t1  .. ..  . .   n ∂ξ ∂ϕ  0

0

∂tn−1

∂tn−1

has rank n. We will in fact require the stronger assumption that the determinant obtained by deleting the last column be 6= 0, for then, as we shall see, the implicit function theorem gives the solution. The function n+1 f is found as follows. Consider Ft , the flow and let S be the manifold obtained Sof the vector field Z in R by sweeping out Γ by Ft . That is, S = { Ft (Γ) | t ∈ R }. The condition that Z never be tangent to Γ insures that the manifold Γ is “dragged along” by the flow Ft to produce a manifold of dimension n. If S is described by f = f (x1 , . . . , xn ); that is, if S is the graph of f , then f is the solution of the partial differential equation (4.2.17). Indeed, since S is obtained by flowing out by Z, its tangent space contains the vector Z; that is, this vector is perpendicular to the normal of the surface S, namely (∂f /∂x1 , . . . ∂f /∂xn , −1) and thus the PDE (4.2.17) is satisfied. We work parametrically and write a point on S as the flow-out of a point on Γ:
Ft ξ01 (t1 , . . . , tn−1 ), . . . ξ0n (t1 , . . . , tn−1 ), ϕ0 (t1 , . . . , tn−1 ) and we denote this point as xi = ξ i (t1 , . . . , tn−1 , t), i = 1, . . . , n

and f = ϕ(t1 , . . . , tn−1 , t),

so that ξ i (t1 , . . . , tn−1 , 0) = ξ0i (t1 , . . . , tn−1 ) and ϕ(t1 , . . . , tn−1 , 0) = ϕ0 (t1 , . . . , tn−1 ). Assuming that ∂x1 n ∂xn X1 · · · X ··· ∂t ∂t n ∂x1 1 n ∂x ∂x ∂x ··· ∂t ··· ∂t 1 = ∂t1 ∂t 0 6= . 1 1 . . . . . . . . · · · . ··· . . n 1 ∂x ∂x ∂x1 ∂xn ∂t ∂tn−1 n−1 ∂tn−1 ∂tn−1 (evaluated at t = 0), one can locally invert to give t = t(x1 , . . . , xn ), ti = ti (x1 , . . . , xn ), i = 1, . . . , n − 1. Substitution into f yields f = f (x1 , . . . , xn ) and hence an implicitly defined solution.

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The fundamental assumption in this construction is that the characteristic vector field Z is never tangent to the (n − 1)-manifold Γ. The method breaks down if one uses manifolds Γ, which are tangent to this vector field at some point. The reason is that at such a point, no complete information about the derivative of f in a complementary (n − 1)-dimensional subspace to the characteristic is known. 4.2.35 Example.

Consider the PDE on R2 given by ∂f ∂f +f = 3, ∂x ∂y

for an unknown function f (x, y) that satisfies the condition   1 2 f s, s − s = s 2 for each s ∈ R. The given initial condition corresponds s, f = s }. On this one-manifold 1 ∂x ∂s

(4.2.18)

to the curve in R3 given by Γ = { (x, y, f ) | x = s, y = (1/2)s2 − f 1 s = −1 6= 0, ∂y = 1 s − 1 ∂s

so that the vector field ∂/∂x + f ∂/∂y + 3∂/∂f is never tangent to Γ. Its flow is easily computed to be   3 2 Ft (x, y, f ) = t + x, t + f t + y, 3t + f 2 so that the manifold swept out by Γ along Ft is given by x(t, s) = t + s, y(t, s) = (3/2)t2 + st + (1/2)s2 − s, f (t, s) = 3t + s. Eliminating t, s we get p f (x, y) = x − 1 ± 1 − 2x2 + 4x + 4y. Only one of thesepfunctions is consistent with the initial conditions given in equation (4.2.18), namely f (x, y) = x − 1 + 1 − 2x2 + 4x + 4y, which is defined for 1 − 2x2 + 4x + 4y ≥ 0.  Another interesting phenomenon occurs when S can no longer be described in terms of the graph of f ; for example, S “folds over.” This corresponds to the formation of shock waves. Further information can be found in Chorin and Marsden [1993], Lax [1973], Guillemin and Sternberg [1977], John [1975], and Smoller [1983].

Supplement 4.2C Automorphisms of Function Algebras The property of flows corresponding to the derivation property of vector fields is that they are an algebra preserving Ft∗ (f g) = (Ft∗ f )(Ft∗ g). In fact it is obvious that every diffeomorphism induces an algebra automorphism of F(M ). The following theorem shows the converse. (We note that there is an analogous result of Mackey [1962] for measurable functions and measurable automorphisms.)

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4.2.36 Theorem. Let M be a paracompact second–countable finite dimensional manifold. Let α : F(M ) → F(M ) be an invertible linear mapping that satisfies α(f g) = α(f )α(g) for all f, g ∈ F(M ). Then there is a unique C ∞ diffeomorphism ϕ : M → M such that α(f ) = f ◦ ϕ. Remarks. A. There is a similar theorem for paracompact second–countable Banach manifolds; here we assume that there are invertible linear maps αF : C ∞ (M, F ) → C ∞ (M, F ) for each Banach space F such that for any bilinear continuous map B : F × G → H we have αH (B(f, g)) = B(αF (f ), αG (g)) for f ∈ C ∞ (M, F ) and g ∈ C ∞ (M, G). The conclusion is the same: there is a C ∞ diffeomorphism ϕ : M → M such that αF (f ) = f ◦ ϕ for all F and all f ∈ C ∞ (M, G). Alternative to assuming this for all F , one can take F = R and assume that M is modelled on a Banach space that has a norm that is C ∞ away from the origin. We shall make some additional remarks on the infinite-dimensional case in the course of the proof. B. Some of the ideas about partitions of unity are needed in the proof. Although the present proof is self–contained, the reader may wish to consult §5.6 simultaneously. C. In Chapter 5 we shall see that finite-dimensional paracompact manifolds are metrizable, so by Theorem 1.6.14 they are automatically second countable. 

Proof of uniqueness. We shall first construct a C ∞ function χ : M → R which takes on the values 1 and 0 at two given points m1 , m2 ∈ M , m1 6= m2 . Choose a chart (U, ϕ) at m1 , such that m2 6∈ U and such that ϕ(U ) is a ball of radius r1 about the origin in E, ϕ(m1 ) = 0. Let V ⊂ U by the inverse image by ϕ of the ball of radius r2 < r1 and let θ : E → R be a C ∞ -bump function as in Lemma 4.2.13. Then the function χ : M → R given by ( θ ◦ ϕ, on U ; χ= 0, on M \U. is clearly C ∞ and χ(m1 ) = 1, χ(m2 ) = 0. Now assume that ϕ∗ f = ψ ∗ f for all f ∈ F(M ) for two different diffeomorphisms ϕ, ψ of M . Then there is a point m ∈ M such that ϕ(m) 6= ψ(m) and thus we can find χ ∈ F(M ) such that (χ ◦ ϕ)(m) = 1, (χ ◦ ψ)(m) = 0 contradicting ϕ∗ χ = ψ ∗ χ. Hence ϕ = ψ.  The proof of existence is based on the following key lemma. 4.2.37 Lemma. Let M be a (finite-dimensional ) paracompact second countable manifold and β : F(M ) → R be a nonzero algebra homomorphism. Then there is a unique point m ∈ M such that β(f ) = f (m). Proof. (Following suggestions of H. Bercovici.) Uniqueness is clear, as before, since for m1 6= m2 there exists a bump function f ∈ F(M ) satisfying f (m1 ) = 0, f (m2 ) = 1. To show existence, note first that β(1) = 1. Indeed β(1) = β(12 ) = β(1)β(1) so that either β(1) = 0 or β(1) = 1. But β(1) = 0 would imply β is identically zero since β(f ) = β(1 · f ) = β(1) · β(f ), contrary to our hypotheses. Therefore we must have β(1) = 1 and thus β(c) = c for c ∈ R. For m ∈ M , let Ann(m) = { f ∈ F(M ) | f (m) = 0 }. Second, we claim that it is enough to show that there is an m ∈ M such that ker β = { f ∈ F(M ) | β(f ) = 0 } = Ann(m). Clearly, if β(f ) = f (m) for some m, then ker β = Ann(m). Conversely, if this holds for some m ∈ M and f 6∈ ker β, let c = β(f ) and note that f − c ∈ ker β = Ann(m), so f (m) = c and thus β(f ) = f (m) for all f ∈ F(M ).

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To prove that ker β = Ann(m) for some m ∈ M , note that both are ideals in F(M ); that is, if f ∈ ker β, (resp., Ann(m)), and g ∈ F(M ), then f g ∈ ker β (resp., Ann(m)). Moreover, both of them are maximal ideals; that is, if I is another ideal of F(M ), with I 6= F(M ), and ker β ⊂ I, (resp., Ann(m) ⊂ I) then necessarily ker β = I (resp. Ann(m) = I). For ker β this is seen in the following way: since R is a field, it has no ideals except 0 and itself; but β(I) is an ideal in R, so β(I) = 0, that is, I = ker β, or β(I) = R = β(F(M )). If β(I) = R, then for every f ∈ F(M ) there exists g ∈ I such that f − g ∈ ker β ⊂ I and hence f ∈ g + I ⊂ I; that is, I = F(M ). Similarly, the ideal Ann(m) is maximal since the quotient F(M )/Ann(m) is isomorphic to R. Assume that ker β 6= Ann(m) for every m ∈ M . By maximality, neither set can be included in the other, and hence for every m ∈ M there is a relatively compact open neighborhood Um of m and fm ∈ ker β such that fm |Um > 0. Let Vm be an open neighborhood of the closure, cl(Um ). Since M is paracompact, we can assume that { Vm | m ∈ M } is locally finite. Since M is second countable, M can be covered by { Vm(j) | j ∈ N }. Let fj = fm(j) and let χj be bump functions which are equal to 1 on cl(Um(j) ) and vanishing in M \Vm(j) . If we have the inequality an <

n2

1 , sup{ χn (m)fn2 (m) | m ∈ M }

then the function f=

X

an χn fn2

n≥1 ∞

is C (since the sum is finite in a neighborhood of every point), f > 0 on M , and the series defining f is P uniformly convergent, being majorized by n≥1 n−2 . If we can show that β can be taken inside the sum, then β(f ) = 0. This construction then produces f ∈ ker β, f > 0 and hence 1 = (1/f )f ∈ ker β; that is, ker β = F(M ), contradicting the hypothesis β 6= 0. To show that β can be taken inside the series, it suffices to prove the following “g-estimate”: for any g ∈ F(M ), |β(g)| ≤ sup{ |g(m)| | m ∈ M }. Once this is done, then N X 2 β(an χn fn ) − β(f ) = β m=1

! −f m=1 N X ≤ sup an χn fn2 − f → 0 N X

an χn fn2

m=1

P as n → ∞ by uniform convergence and boundedness of all functions involved. Thus β(f ) = n≥1 β(an χn fn2 ). To prove the g-estimate, let λ > sup{ |g(m)| | m ∈ M } so that λ ± g 6= 0 on M ; that is, λ ± g are both invertible functions on M . Since β is an algebra homomorphism, 0 6= β(λ ± g) = λ ± β(g). Thus ±β(g) 6= λ for all λ > sup{ |g(m)| | m ∈ M }. Hence we get |β(g)| ≤ sup{ |g(m)| | m ∈ M }, which is the required estimate, so the proof is complete.



We note that for infinite-dimensional manifolds, the proof of the lemma is almost identical, with the following changes: we work with β : C ∞ (M, F ) → F , absolute values are replaced by norms, second countability is in the hypothesis, and the neighborhoods Vm are chosen in such a way that fm |Vm is a bounded function (which is possible by continuity of fm ).

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Proof of existence in Theorem 4.2.36. For each m ∈ M , define the algebra homomorphism βm : F(M ) → R by βm (f ) = α(f )(m). Since α is invertible, α(1) 6= 0 and since α(1) = α(12 ) = α(1)α(1), we have α(1) = 1. Thus βm 6= 0 for all m ∈ M . By Lemma 4.2.37 there exists a unique point, which we call ϕ(m) ∈ M , such that βm (f ) = f (ϕ(m)) = (ϕ∗ f )(m). This defines a map ϕ : M → M such that α(f ) = ϕ∗ f for all f ∈ F(M ). Since α is an automorphism, ϕ is bijective and since α(f ) = ϕ∗ f ∈ F(M ), α−1 (f ) = ϕ∗ f ∈ F(M ) for all f ∈ F(M ), both ϕ, ϕ−1 are C ∞ (take for f any coordinate function multiplied by a bump function to show that in every chart the local representatives of ϕ, ϕ−1 are C ∞ ).  The proof of existence in the infinite-dimensional case proceeds in a similar way.

Supplement 4.2D Derivations on C r Functions This supplement investigates to what extent vector fields and tangent vectors are characterized by their derivation properties on functions, if the underlying manifold is finite dimensional and of a finite differentiability class. We start by studying vector fields. Recall from Proposition 4.2.9 that a derivation θ is an R-linear map from F k+1 (M ) to F k (M ) satisfying the Leibniz rule, that is, θ(f g) = f θ(g) + gθ(f ) for f, g ∈ F k+1 (M ), if the differentiability class of M is at least k + 1. 4.2.38 Theorem (A. Blass). Let M be a C k+2 finite-dimensional manifold, where k ≥ 0. The collection of all derivations θ from F k+1 (M ) to F k (M ) is isomorphic to Xk (M ) as a real vector space. Proof. By the remark following Theorem 4.2.16, there is a unique C k vector field X with the property that θ|F k+2 (M ) = £X |F k+2 (M ). Thus, all we have to do is show that θ and £X agree on C k+1 functions. Replacing θ with θ − £X , we can assume that θ annihilates all C k+2 functions and we want to show that it also annihilates all C k+1 functions. As in the proof of Theorem 4.2.16, it suffices to work in a chart, so we can assume without loss of generality that M = Rn . Let f be a C k+1 function and fix p ∈ Rn . We need to prove that (θf )(p) = 0. For simplicity, we will show this for p = 0, the proof for general p following by centering the following arguments at p instead of 0. Replacing f by the difference between f and its Taylor polynomial of order k + 1 about 0, we can assume f (0) = 0, and the first k + 1 derivatives vanish at 0, since θ evaluated at the origin annihilates any k+1 polynomial. We shall prove that f = g + h, where g and functions, satisfying g|U = 0 and T h are two C h|V = 0, where U, V are open sets such that 0 ∈ cl(U ) cl(V ). Then, since θ is a local operator, θg|U = 0 and θh|V = 0, whence by continuity θg|cl(U ) = 0 and θh|cl(V ) = 0. Hence (θf )(0) = (θg)(0) + (θh)(0) = 0 and the theorem will be proved. 4.2.39 Lemma. Let ϕ : S n−1 ⊂ Rn → R be a C ∞ function and denote by π : Rn \{0} → S n−1 , π(x) = x/kxk the radial projection. Then for any positive integer r, Dr (ϕ ◦ π)(x) =

(ψ ◦ π)(x) kxkr

for some C ∞ function ψ : S n−1 → Lrs (Rn ; R). In particular Dk (ϕ ◦ π)(x) = O(kxk−r ) as kxk → 0. Proof.

For r = 0 choose ϕ = ψ. For r = 1, note that Dπ(x) · v = −

1 v Dk · k(x) · v + , 2 kxk kxk

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and so   1 1 D(ϕ ◦ π)(x) · v = Dϕ(π(x)) I − Dk · k(x) · v. kxk kxk But the mapping `0 (x) = (1/kxk)Dk·k(x) satisfies `0 (tx) = `0 (x) for all t > 0 so that it is uniquely determined by ` = `0 |S n−1 . Hence 1 (ψ ◦ π)(x), D(ϕ ◦ π)(x) = kxk where ψ(y) = Dϕ(y) · (I − `(y)), y ∈ S n−1 . Now proceed by induction.

H

Returning to the proof of the theorem, let f be as before, that is, of class C k+1 and Di f (0) = 0, 0 ≤ i ≤ k + 1, and let ϕ, π be as in the lemma. From Taylor’s formula with remainder, we see that Di f (x) = o(kxkk+1−i ), 0 ≤ i ≤ k + 1, as x → 0. Hence by the product rule and the lemma, Di (f · (ϕ ◦ π))(x) =

X

o(kxkk+1−j )O(kxk−` ) = o(kxkk+1−i )

j+`=i

so that Di (f · (ϕ ◦ π)), 0 ≤ i ≤ k + 1, can be continuously extended to 0, by making them vanish at 0. Therefore f · (g ◦ π) is C k+1 for all Rn . Now choose the C ∞ function ϕ in the lemma to be zero on an open set O1 and equal to 1 on an open set O2 of S n−1 , O1 ∩ O2 = ∅. Then the continuous extension g of f · (ϕ ◦ π) to Rn is zero on U = π −1 (O1 ) and agrees with f on V = π −1 (O2 ). Let h = f − g and thus f is the sum of two C k+1 functions, each of which vanishes in an open set having 0 in its closure. This completes the proof.  We do not know of an example of a derivation not given by a vector field on a C 1 -manifold. In infinite dimensions, the proof would require the norm of the model space to be C ∞ away from the origin and the function ψ in the lemma bounded with all derivatives bounded on the unit sphere. Unfortunately, this does not seem feasible under realistic hypotheses. The foregoing proof is related to the method of “blowing-up” a singularity; see for example Takens [1974] and Buchner, Marsden, and Schecter [1983a]. There are also difficulties with this method in infinite dimensions in other problems, such as the Morse lemma (see Golubitsky and Marsden [1983] and Buchner, Marsden, and Schecter [1983b]). 4.2.40 Corollary. Let M be a C k+1 finite-dimensional manifold. Then the only derivative from F k+1 (M ) to F k (M ), where l ≤ k < ∞, is zero. Proof. By the theorem, such a derivation is given by a C k−1 vector field X. If X 6= 0, then for some f ∈ F k+1 (M ), X[f ] is only C k−1 but not C k .  Next, we turn to the study of the relationship between tangent vectors and germ derivations. On F k (M ) consider the following equivalence relation: f ∼m g iff f and g agree on some neighborhood of m ∈ M . k Equivalence classes of the relation ∼m are called germs at m; they form a vector space denoted by Fm (M ). k The differential d on functions clearly induces an R-linear map, denoted by dm on Fm (M ) by dm f = df (m), k ∗ where we understand f on the left hand side as a germ. It is straightforward to see that dm : Fm (M ) → Tm M k is R-linear and satisfies the Leibniz rule. We say that an R-linear map θm : Fm (M ) → E, where E is a ∗ Banach space, is a germ derivation if θm satisfies the Leibniz rule. Thus, dm is a Tm M -valued germ derivation. Any tangent vector vm ∈ Tm M defines an R-valued germ derivation by vm [f ] = hdf (m), vm i. Conversely, localizing the statement and proof of Theorem 4.2.16(i) at m, we see that on a C ∞ finite-dimensional manifold, any R-valued germ derivation at m defines a unique tangent vector, that is, Tm M is isomorphic to the vector space of R-valued germ derivations on Fm (M ). The purpose of the rest of this supplement is to show that this result is false if M is a C k -manifold. This is in sharp contrast to Theorem 4.2.38.

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259

4.2.41 Theorem ( Newns and Walker [1956]). Let M be a finite dimensional C k manifold, l ≤ k < ∞. k Then there are R-valued germ derivations on Fm (M ) which are not tangent vectors. In fact, the vector space k of all R-valued germ derivations on Fm (M ) is card(R)-dimensional, where card(R) is the cardinality of the continuum. For the proof, we start with algebraic characterizations of Tm M and the vector space of all germ derivations. 4.2.42 Lemma.

Let k k Fm,0 (M ) = { f ∈ Fm (M ) | df (m) = 0 }.

Then k k ∗ Fm (M )/Fm,0 (M ) is isomorphic to Tm M.

Therefore, since M is finite dimensional k k (Fm (M )/Fm,0 (M ))∗ is isomorphic to Tm M. k k ∗ Proof. The isomorphism of Fm (M )/Fm,0 (M ) with Tm M is given by class of (f ) 7→ df (m); this is a direct consequence of Corollary 4.2.14. H k k (M ), f (m) = g(m) = 0}. Then the space of 4.2.43 Lemma. Let Fm,d (M ) = span{1, f g ⊥ f, g ∈ Fm k k k (M ))∗ . R-linear germ derivations on Fm (M ) is isomorphic to (Fm (M )/Fm,d k (M ) with Proof. Clearly, if θm is a germ derivation θm (1) = 0 and θm (f g) = 0 for any f, g ∈ Fm k k (M ). f (m) = g(m) = 0, so that θm defines a linear functional on Fm (M ) which vanishes on the space Fm,d k k Conversely, if λ is a linear functional on Fm (M ) vanishing on Fm,d (M ), then λ is a germ derivation, for if k (M ), we have f, g ∈ Fm

f g = (f − f (m))(g − g(m)) + f (m)g + g(m)f − f (m)g(m) so that λ(f g) = f (m)λ(g) + g(m)λ(f ), that is, the Leibniz rule holds. 4.2.44 Lemma.

H

k All germs in Fm,d (M ) have k + 1 derivatives at m (even though M is only of class C k ).

k k Proof. Since any element of Fm,d (M ) is of the form a + bf g, f, g ∈ Fm (M ), f (m) = g(m) = 0, a, b ∈ R, it suffices to prove the statement for f g. Passing to local charts, we have by the Leibniz rule

Dk (f g) = (Dk f )g + f (Dk g) + ϕ, for ϕ=

k−1 X

(Di f )(Dk−i g).

i=1 1

Clearly ϕ is C , since the highest order derivative in the expression of ϕ is k − 1 and f, g are C k . Moreover, since Dk f is continuous and g(m) = 0, using the definition of the derivative it follows that D[(Dk f )g](m) = (Dk f )(m)(Dg)(m). Therefore, f g has k + 1 derivatives at m. H Proof of Theorem 4.2.41.

We clearly have k k Fm,d (M ) ⊂ Fm,0 (M ).

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Choose a chart (x1 , . . . , xn ) at m, xi (m) = 0, and consider the functions |xi |k+ε , 0 < ε < 1. These functions k k are clearly in Fm,0 (M ), but are not in Fm,d (M ) by Lemma 4.2.44, since they cannot be differentiated k + 1 k k times at m. Therefore, Fm,d (M ) is strictly contained in Fm,0 (M ) and thus Tm M is a strict subspace of the k vector space of germ differentiations on Fm (M ) by Lemmas 4.2.42 and 4.2.43. k k The functions |xi |k+ε , 0 < ε < 1 are linearly independent in Fm M modulo Fm,d (M ), because only a trivial linear combination of such functions has derivatives of order k + 1 at m. Therefore, the dimension of k k k k k Fm (M )/Fm,d (M ) is at least card(R). Since card(Fm (M )) = card(R), it follows that dim(Fm (M )/Fm,d (M )) = card(R). Consequently, its dual, which by Lemma 4.2.43 coincides with the vector space of germ-derivations at m, also has dimension card(R). 

Exercises  4.2-1.

Compute the brackets [X, Y ] for the following vector fields:

(i) In the plane, X = (x + y)

∂ ∂ − ; ∂x ∂y

Y =y

∂ ∂ +x ∂x ∂y

(ii) In space, X=x  4.2-2.

∂ ∂ ∂ +y +z ; ∂x ∂y ∂z

Y = yz

∂ ∂ ∂ + xz + xy ∂x ∂y ∂z

(i) On R2 , let X(x, y) = (x, y; y, −x). Find the flow of X.

(ii) Solve the following for f (t, x, y): ∂f ∂f ∂f =y −x ∂t ∂x ∂y if f (0, x, y) = y sin x.  4.2-3. (i) Let X and Y be vector fields on M with complete flows Ft and Gt , respectively. If [X, Y ] = 0, show that X + Y has flow Ht = Ft ◦ Gt . Is this true if X and Y are time dependent? (ii) Show that if [X, Y ] = 0 for all Y ∈ X(M ), then X = 0. Hint: From the local formula conclude first that X is constant; then take for Y linear vector fields and apply the Hahn–Banach theorem. In infinite dimensions, assume the conditions hold locally or that the model spaces are C ∞ .  4.2-4. Show that, under suitable hypotheses, that the solution f (x, t) = g(Ft (x)) of problem (P) given in Proposition 4.2.11 is unique. Hint: Consider the function Z E(t) = |f1 (x, t) − f2 (x, t)|2 dx Rn where f1 and f2 are two solutions. Show that dE/dt ≤ αE for a suitable constant α and conclude by Gronwall’s inequality that E = 0. The “suitable hypotheses” are conditions that enable integration by parts to be performed in the computation of dE/dt. Adapt this proof to get uniqueness of the solution in Proposition 4.2.28.  4.2-5.

Let X, Y ∈ X(M ) have flows Ft and Gt , respectively. Show that d d [X, Y ] = (F−t ◦ Gs ◦ Ft ). dt ds t,s=0

Hint: The flow of Ft∗ Y is s 7→ F−t ◦ Gs ◦ Ft .

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261

 4.2-6. Show that SO(n) is parallelizable. See Exercise 3.5-19 for a proof that SO(n) is a manifold. Hint: SO(n) is a group.  4.2-7.

Solve the following system of partial differential equations. ∂Y 1 ∂Y 1 ∂Y 1 = (x + y) + (4x − 2y) − Y 1 − Y 2, ∂t ∂x ∂y ∂Y 2 ∂Y 2 ∂Y 2 = (x + y) + (4x − 2y) − 4Y 1 + 2Y 2 , ∂t ∂x ∂y

with initial conditions Y 1 (x, y, 0) = x + y, Y 2 (x, y, 0) = x2 . Hint: The flow of the vector field (x + y, 4x − 2y) is   1 1 4 1 (x, y) 7→ (x − y)e−3t + (4x + y)e2t , − (x − y)e−3t + (4x + y)e2t . 5 5 5 5  4.2-8.

Consider the following equation for f (x, t) in divergence form: ∂ ∂f + (H(f )) = 0 ∂t ∂x

where H is a given function of f . Show that the characteristics are given by x˙ = −H 0 (f ). What does the transversality condition discussed in Supplement 4.2B become in this case?  4.2-9. Let M and N be manifolds with N modeled on a Banach space which has a C k norm away from the origin. Show that a given mapping ϕ : M → N is C k iff f ◦ ϕ : M → R is C k for all f ∈ F k (N ).  4.2-10. Develop a product formula like that in Supplement 4.1A for the flow of X + Y for time-dependent vector fields. Hint: You will have to consider time-ordered products.  4.2-11 (Newns and Walker [1956]). In the terminology of Supplement 4.2D, consider a C 0 -manifold modeled on Rn . Show that any germ derivation is identically zero. 0 (M ), f = f (m) + (f − f (m))1/3 (f − f (m))2/3 and apply the derivation. Hint: Write any f ∈ Fm  4.2-12 (More on the Lie bracket as a “commutator”). Let M be a manifold, m ∈ M , v ∈ Tm M . Recall that Tm M is a submanifold of T (T M ) and that Tv (Tm M ) is canonically isomorphic to Tm M . Also recall from Exercise 3.3-2 that on T (T M ) there is a canonical involution sM : T (T M ) → T (T M ) satisfying sM ◦ sM = identity on T (T M ), T τM ◦ sM = τT M , and τT M ◦ sM = T τM , where τM : T M → M and τT M : T (T M ) → T M are the canonical tangent bundle projections. Let X, Y ∈ X(M ) and denote by T X, T Y : T M → T (T M ) their tangent maps. Prove the following formulae for the Lie bracket: [X, Y ](m) = sM (Tm Y (X(m))) − Tm X(Y (m)) = Tm Y (X(m)) − sM (Tm X(Y (m))), where the right hand sides, belonging to TX(m) (Tm M ) and TY (m) (Tm M ) respectively, are thought of as elements of Tm M . Hint: Show that Tm τM of the right hand sides is zero which proves that the right hand sides are not just elements of TX(m) (T M ) and TY (m) (T M ) respectively, but of the indicated spaces. Then pass to a local chart.

4.3

An Introduction to Dynamical Systems

We have seen quite a bit of theoretical development concerning the interplay between the two aspects of vector fields, namely as differential operators and as ordinary differential equations. It is appropriate now to

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look a little more closely at geometric aspects of flows. Much of the work in this section holds for infinitedimensional as well as finite-dimensional manifolds. The reader who knows or is willing to learn some spectral theory from functional analysis can make the generalization. This section is intended to link up the theory of this book with courses in ordinary differential equations that the reader may have taken. The section will be most beneficial if it is read with such a course in mind. Equilibrium Points. and their stability.

We begin by introducing some of the most basic terminology regarding fixed points

4.3.1 Definition. Let X be a C 1 vector field on an n-manifold M . A point m is called a critical point (also called a singular point or an equilibrium point) of X if X(m) = 0. The linearization of X at a critical point m is the linear map X 0 (m) : Tm M → Tm M defined by d (T Ft (m) · v) X (x) · v = dt t=0 0

where Ft is the flow of X. The eigenvalues (points in the spectrum) of X 0 (m) are called characteristic exponents of X at m. Some remarks will clarify this definition. Ft leaves m fixed: Ft (m) = m, since c(t) ≡ m is the unique integral curve through m. Conversely, it is obvious that if Ft (m) = m for all t, then m is a critical point. Thus Tm Ft is a linear map of Tm M to itself and so its t-derivative at 0, producing another linear map of Tm M to itself, makes sense. 4.3.2 Proposition. Let m be a critical point of X and let (U, ϕ) be a chart on M with ϕ(m) = x0 ∈ Rn . Let x = (x1 , . . . , xn ) denote coordinates in Rn and X i (x1 , . . . , xn ), i = 1, . . . , n, the components of the local representative of X. Then the matrix of X 0 (m) in these coordinates is   ∂X i . ∂xj x=x0 Proof.

This follows from the equations X i (Ft (x)) =

d i F (x) dt t

after differentiating in x and setting x = x0 , t = 0.



The name “characteristic exponent” arises as follows. We have the linear differential equation d Tm Ft = X 0 (x) ◦ Tm Ft dt and so Tm Ft = etX

0

(m)

.

Here the exponential is defined, for example, by a power series. The actual computation of these exponentials is learned in differential equations courses, using the Jordan canonical form. (See Hirsch and Smale [1974], for instance.) In particular, if µ1 , . . . , µn are the characteristic exponents of X at m, the eigenvalues of Tm Ft are etµ1 , . . . , etµn . The characteristic exponents will be related to the following notion of stability of a critical point. 4.3.3 Definition.

Let m0 be a critical point of X. Then

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263

(i) m0 is stable (or Liapunov stable) if for any neighborhood U of m0 , there is a neighborhood V of m0 such that if m ∈ V , then m is + complete and Ft (m) ∈ U for all t ≥ 0. (See Figure 4.3.1(a).) (ii) m0 is asymptotically stable if there is a neighborhood V of m0 such that if m ∈ V , then m is + complete, Ft (V ) ⊂ Fs (V ) if t > s and lim Ft (V ) = {m0 },

t→+∞

that is, for any neighborhood U of m0 , there is a T such that Ft (V ) ⊂ U if 4.3.1(b).)

t ≥ T . (See Figure

It is obvious that asymptotic stability implies stability. The harmonic oscillator x ¨ = −x giving a flow in the plane shows that stability need not imply asymptotic stability (Figure 4.3.1(c)). y U V V m0

m x

m0

m

Figure 4.3.1. Stability of equilibria.

4.3.4 Theorem (Liapunov’s Stability Criterion). Suppose X is C 1 and m is a critical point of X. Assume the spectrum of X 0 (m) is strictly in the left half plane. (In finite dimensions, the characteristic exponents of m have negative real parts.) Then m is asymptotically stable. (In a similar way, if Re(µi ) > 0, m is asymptotically unstable, that is, asymptotically stable as t → −∞.) The proof we give requires some spectral theory that we shall now review. For the finite dimensional case, consult the exercises. This proof in fact can be adapted to work for many partial differential equations (see Marsden and Hughes [1983, Chapters 6, 7, and p. 483]). Let T : E → E be a bounded linear operator on a Banach space E and let σ(T ) denote its spectrum; that is, σ(T ) = { λ ∈ C | T − λI is not invertible on the complexification of E }. Then σ(T ) is non-empty, is compact, and for λ ∈ σ(T ), |λ| ≤ kT k. Let r(T ) denote its spectral radius, defined by r(T ) = sup{ |λ| | λ ∈ σ(T ) }. 4.3.5 Theorem (Spectral Radius Formula). r(t) = lim kT n k1/n . n→∞

The proof is analogous to the formula for the radius of convergence of a power series and can be supplied without difficulty; cf. Rudin [1973, p. 355]. The following lemma is also not difficult and is proved in Rudin [1973] and Dunford and Schwartz [1963]. 4.3.6 Theorem (Spectral Mapping Theorem).

Let

f (z) =

∞ X n=0

an z n

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be an entire function and define ∞ X

f (T ) =

an T n .

n=0

Then σ(f (T )) = f (σ(T )). 4.3.7 Lemma. Let T : E → E be a bounded linear operator on a Banach space E. Let r be any number greater than r(T ), the spectral radius of T . Then there is a norm | · | on E equivalent to the original norm such that |T | ≤ r. Proof.

From the spectral radius formula, we get supn≥0 (kT n k/rn ) < ∞, so if we define kT n (x)k , rn n≥0

|x| = sup then | | is a norm and  kxk ≤ |x| ≤ Hence

kT n k sup n n≥0 r

 kxk.

kT n+1 (x)k kT n+1 (x)k = r sup ≤ r|x|. n r rn+1 n≥0 n≥0

|T (x)| = sup which proves the result.

(4.3.1) 

4.3.8 Lemma. Let A : E → E be a bounded operator on E and let r > σ(A) (i.e., if λ ∈ σ(A), Re(λ) > r). Then there is an equivalent norm | · | on E such that for t ≥ 0, |etA | ≤ ert . Proof.

Using Theorem 4.3.6, ert is ≥ spectral radius of etA ; that is, ert ≥ limn→∞ kentA k1/n . Set |x| =

kentA (x)k ernt n≥0,t≥0 sup

and proceed as in Lemma 4.3.7.



Write r bigger than spectral radius? Check. State something about using bounded operators in the introduction to the section. There is an analogous lemma if r < σ(A), giving |etA | ≥ ert . 4.3.9 Lemma. Let T : E → E be a bounded linear operator. Let σ(T ) ⊂ { z | Re(z) < 0 } (resp., σ(T ) ⊂ { z | Re(z) > 0 }). Then the origin is an attracting (resp., repelling) fixed point for the flow ϕt = etT of T . Proof. If σ(T ) ⊂ { z | Re(z) < 0 }, there is an r < 0 with σ(T ) < r, as σ(T ) is compact. Thus by Lemma 4.3.8, |etA | ≤ ert → 0 as t → +∞.  Proof of Liapunov’s Stability Criterion Theorem 4.3.4. Without loss of generality, we can assume that M is a Banach space E and that m = 0. As above, renorm E and find ε > 0 such that ketA k ≤ e−εt , where A = X 0 (0).

ToDo

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From the local existence theory, there is an r-ball about 0 for which the time of existence is uniform if the initial condition x0 lies in this ball. Let R(x) = X(x) − DX(0) · x. Find r2 ≤ r such that kxk ≤ r2 implies kR(x)k ≤ αkxk, where α = ε/2. Let D be the open r2 /2 ball about 0. We shall show that if x0 ∈ D, then the integral curve starting at x0 remains in D and → 0 exponentially as t → +∞. This will prove the result. Let x(t) be the integral curve of X starting at x0 . Suppose x(t) remains in D for 0 ≤ t < T . The equation Z x(t) = x0 +

t

Z X(x(s)) ds = x0 +

0

t

[Ax(s) + R(x(s))] ds 0

gives, by the variation of constants formula (Exercise 4.1-5), x(t) = etA x0 +

Z

t

e(t−s)A R(x(s)) ds,

0

and so kx(t)k ≤ e−tε kx0 k + α

t

Z

e−(t−s)ε kx(s)k ds.

0

Letting f (t) = etε kx(t)k, the previous inequality becomes Z f (t) ≤ kx0 k + α

1

f (s) ds, 0

and so, by Gronwall’s inequality, f (t) ≤ kx0 keαt . Thus kx(t)k ≤ kx0 ke(α−ε)t = kx0 ke−εt/2 . Hence x(t) ∈ D, 0 ≤ t < T , so as in Proposition 4.1.19, x(t) may be indefinitely extended in t and the foregoing estimate holds.  One can also show that if M is finite dimensional and m is a stable equilibrium, then no eigenvalue of X 0 (m) has strictly positive real part; see Hirsch and Smale [1974, pp. 187–190] and the remarks below on invariant manifolds. See Hille and Phillips [1957], and Curtain and Pritchard [1977] for the infinite dimensional linear case. Liapunov Functions.

Another method of proving stability is to use Liapunov functions.

4.3.10 Definition. Let X ∈ Xr (M ), r ≥ 1, and let m be an equilibrium solution for X, that is, X(m) = 0. A Liapunov function for X at m is a continuous function L : U → R defined on a neighborhood U of m, differentiable on U \{m}, and satisfying the following conditions: (i) L(m) = 0 and L(m0 ) > 0 if m0 6= m; (ii) X[L] ≤ 0 on U \{m}; (iii) there is a connected chart ϕ : V → E where m ∈ V ⊂ U , ϕ(m) = 0, and an ε > 0 satisfying Bε (0) = { x ∈ E | kxk ≤ ε } ⊂ ϕ(V ), such that for all 0 < ε0 ≤ ε, inf{ L(ϕ−1 (x)) | kxk = ε0 } > 0. The Liapunov function L is said to be strict, if (ii) is replaced by (ii)0 X[L] < 0 in U \{m}.

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Conditions (i) and (iii) are called the potential well hypothesis. In finite dimensions, (iii) follows automatically from compactness of the sphere of radius ε0 and (i). By the Lie derivative formula, condition (ii) is equivalent to the statement: L is decreasing along integral curves of X. 4.3.11 Theorem. Let X ∈ Xr (M ), r ≥ 1, and m be an equilibrium of X. If there exists a Liapunov function for X at m, then m is stable. Proof. Since the statement is local, we can assume M is a Banach space E and m = 0. By Lemma 4.1.8, there is a neighborhood U of 0 in E such that all solutions starting in U exist for time t ∈ [−δ, δ], with δ depending only on X and U , but not on the solution. Now fix ε > 0 as in (iii) such that the open ball Dε (0) is included in U . Let ρ(ε) > 0 be the minimum value of L on the sphere of radius ε, and define the open set U 0 = { x ∈ Dε (0) | L(x) < ρ(ε) }. By (i), U 0 6= ∅, 0 ∈ U 0 , and by (ii), no solution starting in U 0 can meet the sphere of radius ε (since L is decreasing on integral curves of X). Thus all solutions starting in U 0 never leave Dε (0) ⊂ U and therefore by uniformity of time of existence, these solutions can be extended indefinitely in steps of δ time units. This shows 0 is stable.  Note that if E is finite dimensional, the proof can be done without appeal to Lemma 4.1.8: since the closed ε-ball is compact, solutions starting in U 0 exist for all time by Proposition 4.1.19. 4.3.12 Theorem. Let X ∈ Xr (M ), r ≥ 1, m be an equilibrium of X, and L a strict Liapunov function for X at m. Then m is asymptotically stable if any one of the following conditions hold: (i) M is finite dimensional ; (ii) solutions starting near m stay in a compact set (i.e., trajectories are precompact); (iii) in a chart ϕ : V → E satisfying (iii) in Definition 4.3.10 the following inequality is valid for some constant a > 0 X[L](x) ≤ −akX(x)k. Proof. We can assume M = E, and m = 0. By Theorem 4.3.11, 0 is stable, so if tn is an increasing sequence, tn → ∞, and x(t) is an integral curve of X starting in U 0 (see the proof of Theorem 4.3.11), the sequence {x(tn )} in E has a convergent subsequence in cases (i) and (ii). Let us show that under hypothesis (iii), {x(tn )} is Cauchy, so by completeness of E it is convergent. For t > s, the inequality Z L(x(t)) − L(x(s)) =

t

Z

t

X[L](x(λ)) dλ ≤ −a s

kX(x(λ))k < 0, s

implies that Z

t

L(x(s)) − L(x(t)) ≥ a s

Z

≥ a

s

kX(x(λ))k dλ

t

X(x(λ)) dλ

= akx(t) − x(s)k, which together with the continuity of λ 7→ L(x(λ)) shows that {x(tn )} is a Cauchy sequence. Thus, in all three cases, there is a sequence tn → +∞ such that x(tn ) → x0 ∈ Dε (0), Dε (0) being given in the proof of Theorem 4.3.11. We shall prove that x0 = 0. Since L(x(t)) is a strictly decreasing function of t by (ii)0 , L(x(t)) > L(x0 ) for all t > 0. If x0 6= 0, let c(t) be the solution of X starting at x0 , so that L(c(t)) < L(x0 ), again since t 7→ L(x(t)) is strictly decreasing. Thus, for any solution c˜(t) starting close to x0 , L(˜ c(t)) < L(x0 ) by continuity of L. Now take c˜(0) = x(tn ) for n large to get the contradiction L(x(tn +t)) < L(x0 ). Therefore x0 = 0 is the only limit point of { x(t) | t ≥ 0 } if x(0) ∈ U 0 , that is, 0 is asymptotically stable. 

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The method of Theorem 4.3.12 can be used to detect the instability of equilibrium solutions. 4.3.13 Theorem. Let m be an equilibrium point of X ∈ Xr (M ), r ≥ 1. Assume there is a continuous function L : U → M defined in a neighborhood of U of m, which is differentiable on U \{m}, and satisfies L(m) = 0, X[L] ≥ a > 0 (respectively, ≤ a < 0) on U \{m}. If there exists a sequence mk → m such that L(mk ) > 0 (respectively, < 0), then m is unstable. Proof. We need to show that there is a neighborhood W of m such that for any neighborhood V of m, V ⊂ U , there is a point mV whose integral curve leaves W . Since m is an equilibrium, by Corollary 4.1.25, there is a neighborhood W1 ⊂ U of m such that each integral curve starting in W1 exists for time at least 1/a. Let W = { m ∈ W1 | L(m) < 1/2 }. We can assume M = E, and m = 0. Let cn (t) denote the integral curve of X with initial condition mn ∈ W . Then Z t L(cn (t)) − L(mn ) = X[L](cn (λ)) dλ ≥ at 0

so that L(cn (1/a)) ≥ 1 + L(mn ) > 1, that is, cn (1/a) 6∈ W . Thus all integral curves starting at the points mn ∈ W leave W after time at most 1/a. Since mn → 0, the origin is unstable.  Note that if M is finite dimensional, the condition X[L] ≥ a > 0 can be replaced by the condition X[L] > 0; this follows, as usual, by local compactness of M . The LaSalle Invariance Principle. A key ingredient in proving more general global asymptotic stability results is the LaSalle invariance principle. It allows one to prove attractivity of more general invariant sets than equilibrium points. 4.3.14 Theorem. Consider the smooth dynamical system on an n-manifold M given by x˙ = X(x) and let Ω be a compact set in M that is (positively) invariant under the flow of X. Let V : Ω → R, V ≥ 0, be a C 1 function such that ∂V ·X ≤0 V˙ (x) = ∂x in Ω. Let S be the largest invariant set in Ω where V˙ (x) = 0. Then every solution with initial point in Ω tends asymptotically to S as t → ∞. In particular, if S is an isolated equilibrium, it is asymptotically stable. The invariance principle is due to Barbashin and Krasovskii [1952], Lasalle and Lefschetz [1961], and Krasovskii [1963]. The book of Khalil [1992] has a nice treatment. In the statement of the theorem, V (x) need not be positive definite, but rather only semidefinite, and that if in particular S is an equilibrium point, the theorem proves that the equilibrium is asymptotically stable. The set Ω in the LaSalle theorem also gives us an estimate of the region of attraction of an equilibrium. This is one of the reasons that this is a more attractive methodology than that of spectral stability tests, which could in principle give a very small region of attraction. In many basic infinite dimensional examples, some technical sharpening of the preceding ideas is necessary for them to be applicable. We refer the reader to LaSalle [1976], Marsden and Hughes [1983, Section 6.6], Hale, Magalhaes and Oliva [1984], and Holm, Marsden, Ratiu, and Weinstein [1985] for more information. 4.3.15 Examples. A.

The vector field X(x, y) = (−y − x5 )

∂ ∂ + (x − 2y 3 ) ∈ X(R2 ) ∂x ∂y

has the origin as an isolated equilibrium. The characteristic exponents of X at (0, 0) are ±i and so Liapunov’s Stability Criterion (Theorem 4.3.4) does not give any information regarding the stability of the origin. If

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we suspect that (0, 0) is asymptotically stable, we can try searching for a Liapunov function of the form L(x, y) = ax2 + by 2 , so we need to determine the coefficients a, b, 6= 0 in such a way that X[L] < 0. We have X[L] = 2ax(−y − x5 ) + 2by(x − 2y 3 ) = 2xy(b − a) − 2ax6 − 4by 4 , so that choosing a = b = 1, we get X[L] = −2(x6 + 2y 4 ) which is strictly negative if (x, y) 6= (0, 0). Thus the origin is asymptotically stable by Theorem 4.3.12. B.

Consider the vector field X(x, y) = (−y + x5 )

∂ ∂ + (x + 2y 3 ) ∂x ∂y

with the origin as an isolated critical point and characteristic exponents ±i. Again Liapunov’s Stability Criterion cannot be applied, so that we search for a function L(x, y) = ax2 + by 2 , a, b 6= 0 in such a way that X[L] has a definite sign. As above we get X[L] = 2ax(−y + x5 ) + 2by(x + 2y 3 ) = 2xy(b − a) + 2ax6 + 4by 4 , so that choosing a = b = 1, it follows that X[L] = 2(x6 + y 4 ) > 0 if (x, y) 6= (0, 0). Thus, by Theorem 4.3.13, the origin is unstable. These two examples show that if the spectrum of X lies on the imaginary axis, the stability nature of the equilibrium is determined by the nonlinear terms. ¨ = −(1/m)∇V (q) written as a first order system q ¨ = v, v ¨ = C. Consider Newton’s equations in R3 , q −(1/m)∇V (q) and so define a vector field X on R3 × R3 . Let (q0 , v0 ) be an equilibrium of this system, so that v0 = 0 and ∇V (q0 ) = 0. In Example 4.1.23B we saw that the total energy E(q, v) =

1 mkvk2 + V (q) 2

is conserved, so we try to use E to construct a Liapunov function L. Since L(q0 , 0) = 0, define L(q, v) = E(q, v) − E(q0 , 0) =

1 mkvk2 + V (q − V (q0 ), 2

which satisfies X[L] = 0 by conservation of energy. If V (q) > V (q0 ) for q 6= q0 , then L is a Liapunov function. Thus we have proved Lagrange’s Stability Theorem: an equilibrium point (q0 , 0) of Newton’s equations for a particle of mass m, moving under the influence of a potential V , which has a local absolute minimum at q0 , is stable. D. Let E be a Banach space and L : E → R be C 2 in a neighborhood of 0. If DL(0) = 0 and there is a constant c > 0 such that D2 L(0)(e, e) > ckek2 for all e, then 0 lies in a potential well for L (i.e., (ii) and (iii) of Definition 4.3.10 hold ). Indeed, by Taylor’s theorem 2.4.15, L(h) − L(0) =

khk2 1 2 D L(0)(h, h) + o(h2 ) ≥ c + o(h2 ). 2 2

Thus, if δ > 0 is such that for all khk < δ, |o(h2 )| ≤ ckhk2 /4, then L(h) − L(0) > ckhk2 /4, that is, inf [L(h) − L(0)] ≥

khk=ε

for ε < δ.

cε 4 

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Unstable Equilibrium Points. A critical point is called hyperbolic or elementary if none of its characteristic exponents has zero real part. A generalization of Liapunov’s theorem called the Hartman– Grobman theorem shows that near a hyperbolic critical point the flow looks like that of its linearization. (See Hartman [1973, Chapter 9] and Nelson [1969, Chapter 3], for proofs and discussions.) In the plane, the possible flows of a linear system      x˙ 1 a b x1 = x˙ 2 c d x2 are summarized in cases (a) through (e) the table below and shown in Figure 4.3.2. To carry out this classification, one makes a change of variables to bring the coefficient matrix into canonical form. (Remember that for real systems, the characteristic exponents occur in conjugate pairs.) Eigenvalues

Real Jordan form   λ1 0 0 λ2

λ1 < 0 < λ2 λ1 < λ2 < 0 λ1 = λ2 < 0

 λ1 = λ2 < 0 λ1 = a + ib, a < 0 λ2 = a − ib, b > 0



λ1 = ib, λ2 = −ib, b>0



0 λ2



a −b b a



0 −b b 0



λ1 1

Name

Part of Fig. 4.3.2 (a) (b) (c)

saddle stable node stable star stable improper node

(d)

stable spiral sink

(e)

center

(f)

If, in case (e), b < 0, then the orientation of the arrows is clockwise instead of counterclockwise, but the spiral is still inward. If, in case (f), b < 0, then the direction of rotation is reversed. If the signs of λ1 , λ2 , are changed in cases (a) to (e), then all arrows in the phase portraits are reversed  and in (b) to (e), “stable”  is λ 0 0 0 replaced by “unstable”. We leave the reader to consider the degenerate cases , and the shear . 0 0 1 0 In the original coordinate system (x1 , x2 ), all phase portraits in Figure 4.3.2 may be rotated and sheared by the change of coordinates.

(a)

(b)

(c)

(d)

(e)

(f)

Figure 4.3.2. Phase portraits for two dimensional equilibria

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Linear Flows on Banach Spaces. theory is required.

To understand the general case of linear flows, a little more spectral

4.3.16 Lemma. Suppose σ(T ) = σ1 ∪σ2 where d(σ1 , σ2 ) > 0. Then there are unique T -invariant subspaces E1 and E2 such that E = E1 ⊕ E2 , σ(Ti ) = σi , where Ti = T |Ei ; Ei is called the generalized eigenspace of σi . The basic idea of the proof is this: let γj be a closed curve with σj in its interior and σk , k 6= j, in its exterior; then Z dz 1 . Tj = 2πi γj zI − T Note that the eigenspace of an eigenvalue λ is not always the same as the generalized eigenspace of λ. In the finite dimensional case, the generalized eigenspace of T is the subspace corresponding to all the Jordan blocks containing λ in the Jordan canonical form. 4.3.17 Lemma.

Let T , σ1 , and σ2 be as in Lemma 4.3.16; assume d(exp(σ1 ), exp(σ2 )) > 0.

Then for the operator exp(tT ), the generalized eigenspace of exp(tTi ) is Ei . Proof.

Write, according to Lemma 4.3.16, E = E1 ⊕ E2 . Thus  ∞ n n ∞  n n X X t T t T tn T n etT (e1 , e2 ) = (e1 , e2 ) = e1 , e2 n! n! n! n=0 n=0 ! ∞ ∞ X tn T n X tn T n e1 , e2 = (etT1 e1 , etT2 e2 ). = n! n! n=0 n=0

From this the result follows easily.



The Nonlinear Case. Now we discuss the generic nonlinear case; that is, let m be a hyperbolic equilibrium of the vector field X and let Ft be its flow. Define the inset of m by In(m) = { m0 ∈ M | Ft (m0 ) → m as t → +∞ } and similarly, the outset is defined by Out(m) = { m0 ∈ M | Ft (m0 ) → m as t → −∞ }. In the case of a linear system, x˙ = Ax, where A has no eigenvalue on the imaginary axis (so the origin is a hyperbolic critical point), In(0) is the generalized eigenspace of the eigenvalues with negative real parts, while Out(0) is the generalized eigenspace corresponding to the eigenvalues with positive real parts. Clearly, these are complementary subspaces. The dimension of the linear subspace In(0) is called the stability index of the critical point. The Hartman–Grobman linearization theorem states that the phase portrait of X near m is topologically conjugate to the phase portrait of the linear system x˙ = Ax, near the origin, where A = X 0 (m), the linearized vector field at m. This means there is a homeomorphism of the two domains, preserving the oriented trajectories of the respective flows. Thus in this nonlinear hyperbolic case, the inset and outset are C 0 submanifolds. Another important theorem of dynamical systems theory, the stable manifold theorem (Smale [1967]) says that in addition, these are smooth, injectively immersed submanifolds, intersecting transversally at the critical point m. See Figure 4.3.3 for an illustration showing part of the inset and outset near the critical point. It follows from these important results that there are (up to topological conjugacy) only a few essentially different phase portraits, near hyperbolic critical points. These are classified by the dimension of their insets, called the stability index , which is denoted by S(X, m) for m an equilibrium, as in the linear case.

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Figure 4.3.3. Insets and outsets

The word index comes up in this context with another meaning. If M is finite dimensional and m is a critical point of a vector field X, the topological index of m is +1 if the number of eigenvalues (counting multiplicities) with negative real part is even and is −1 if it is odd. Let this index be denoted I(X, m), so that I(X, m) = (−1)S(X,m) . The Poincar´ e–Hopf index theorem states that if M is compact and X only has (isolated) hyperbolic critical points, then X

I(X, m) = χ(M )

m is a critical point of X

where χ(M ) is the Euler–Poincar´e characteristic of M . For isolated nonhyperbolic critical points the index is also defined but requires degree theory for its definition—a kind of generalized winding number; see §7.5 or Guillemin and Pollack [1974, p. 133]. We now illustrate these basic concepts about critical points with some classical applications. 4.3.18 Examples. A.

The simple pendulum with linear damping is defined by the second-order equation x ¨ + cx˙ + k sin x = 0

(c > 0).

This is equivalent to the following dynamical system whose phase portrait is shown in Figure 4.3.4: x˙ = v,

v˙ = −cv − k sin x.

The stable focus at the origin represents the motionless, hanging pendulum. The saddle point at (kπ, 0) corresponds to the motionless bob, balanced at the top of its swing.

B.

Another classical equation models the buckling column (see Stoker [1950, Chapter 3, Section 10]): m¨ x + cx˙ + a1 x + a3 x3 = 0

(a1 < 0, a3 , c > 0),

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4. Vector Fields and Dynamical Systems

Figure 4.3.4. The pendulum with linear damping

or equivalently, the planar dynamical system x˙ = v,

v˙ = −

cv a1 x a3 x3 − − m m m

with the phase portrait shown in Figure 4.3.5. This has two stable foci on the horizontal axis, denoted m1 and m2 , corresponding to the column buckling (due to a heavy weight on the top) to either side. The saddle at the origin corresponds to the unstable equilibrium of the straight, unbuckled column.

v

x

Figure 4.3.5. Phase portrait of the buckling column

Note that in this phase portrait, some initial conditions tend toward one stable focus, while some tend toward the other. The two tendencies are divided by the curve, In(0, 0), the inset of the saddle at the origin. This is called the separatrix , as it separates the domain into the two disjoint open sets, In(m0 ) and In(m1 ). The stable foci are called attractors, and their insets are called their basins. See Figure 4.3.6 for the special 3 case x ¨ + x−x+x ˙ = 0. This is a special case of a general theory, which is increasingly important in dynamical

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y

x

Figure 4.3.6. Basins of attraction of x ¨ + x˙ − x + x3 = 0

systems applications. The attractors are regarded as the principal features of the phase portrait; the size of their basins measures the probability of observing the attractor, and the separatrices help find them.  Another basic ingredient in the qualitative theory is the notion of a closed orbit, also called a limit cycle. 4.3.19 Definition. An orbit γ(t) for a vector field X is called closed when γ(t) is not a fixed point and there is a τ > 0 such that γ(t + τ ) = γ(t) for all t. The inset of γ, In(γ), is the set of points m ∈ M such that Ft (m) → γ as t → +∞ (i.e., the distance between Ft (m) and the (compact) set { γ(t) | 0 ≤ t ≤ τ } tends to zero as t → ∞). Likewise, the outset, Out(γ), is the set of points tending to γ as t → −∞. 4.3.20 Example. One of the earliest occurrences of an attractive closed orbit in an important application is found in Baron Rayleigh’s model for the violin string (see Rayleigh [1887, Volume 1, Section 68a]), u ¨ + k1 u˙ + k3 u˙ 3 + ω 2 u = 0,

k1 < 0 < k3 ,

or equivalently, u˙ = v, v˙ = −k1 v − k3 v 3 − ω 2 u, with the phase portrait shown in Figure 4.3.7. This phase portrait has an unstable focus at the origin, with an attractive closed orbit around it. That is, the closed orbit γ is a limiting set for every point in its basin (or inset) In(γ), which is an open set of the domain. In fact the entire plane (excepting the origin) comprises the basin of this closed orbit. Thus every trajectory tends asymptotically to the limit cycle γ and winds around closer and closer to it. Meanwhile this closed orbit is a periodic function of time, in the sense of Definition 4.3.19. Thus the eventual (asymptotic) behavior of every trajectory (other than the unstable constant trajectory at the origin) is periodic; it is an oscillation. This picture thus models the sustained oscillation of the violin string, under the influence of the moving bow. Related systems occur in electrical engineering under the name van der Pol equation. (See Hirsch and Smale [1974, Chapter 10] for a discussion.)  We now proceed toward the analog of Liapunov’s theorem for the stability of closed orbits. To do this we need to introduce Poincar´e maps and characteristic multipliers.

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v

u

Figure 4.3.7. Rayleigh equation

4.3.21 Definition. Let X be a C r vector field on a manifold M , r ≥ 1. A local transversal section of X at m ∈ M is a submanifold S ⊂ M of codimension one with m ∈ S and for all s ∈ S, X(s) is not contained in Ts S. Let X be a C r vector field on a manifold M with C r flow F : DX ⊂ M × R → M , γ a closed orbit of X with period τ , and S a local transversal section of X at m ∈ γ. A Poincar´ e map of γ is a C r mapping Θ : W0 → W1 where: PM1. W0 , W1 ⊂ S are open neighborhoods of m ∈ S, and Θ is a C r diffeomorphism; PM2. there is a C r function δ : W0 → R such that for all s ∈ W0 , (s, τ − δ(s)) ∈ DX , and Θ(s) = F (s, τ − δ(s)); and finally, PM3. if t ∈ [0, τ − δ(s)], then F (s, t) 6∈ W0 (see Figure 4.3.8).

Figure 4.3.8. Poincar´e maps

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4.3.22 Theorem (Existence and Uniqueness of Poincar´e Maps). (i) If X is a C r vector field on M , and γ is a closed orbit of X, then there exists a Poincar´e map of γ. (ii) If Θ : W0 → W1 is a Poincar´e map of γ (in a local transversal section S at m ∈ γ) and Θ0 also (in S 0 at m0 ∈ γ), then Θ and Θ0 are locally conjugate. That is, there are open neighborhoods W2 of m ∈ S, W20 of m0 ∈ S 0 , and a C r -diffeomorphism H : W2 → W20 , such that W2 ⊂ W0 ∩ W1 ,

W20 ⊂ W00 ∩ W10

and the following diagram commutes: Θ−1 (W2 ) ∩ W2

Θ

H ? W20

- W2 ∩ Θ(W2 ) H ? - S0

Θ

Proof. (i) At any point m ∈ γ we have X(m) 6= 0, so there exists a flow box chart (U, ϕ) at m with ϕ(U ) = V ×I ⊂ Rn−1 ×R. Then S = ϕ−1 (V ×{0}) is a local transversal section at m. If F : DX ⊂ M ×R → M is the integral of X, DX is open, so we may suppose U × [−τ, τ ] ⊂ DX , where τ is the period of γ. As Fτ (m) = m ∈ M and Fτ is a homeomorphism, U0 = Fτ−1 (U ) ∩ U is an open neighborhood of m ∈ M with Fτ (U0 ) ⊂ U . Let W0 = S ∩ U0 and W2 = Fτ (W0 ). Then W2 is a local transversal section at m ∈ M and Fτ : W0 → W2 is a diffeomorphism (see Figure 4.3.9).

S

U

Figure 4.3.9. Coordinates adapted to a periodic orbit

If U2 = Fτ (U0 ), then we may regard U0 and U2 as open submanifolds of the vector bundle V × R (by identification using ϕ) and then Fτ : U0 → U2 is a C r diffeomorphism mapping fibers into fibers, as ϕ identifies orbits with fibers, and Fτ preserves orbits. Thus W2 is a section of an open subbundle. More precisely, if π : V × I → V and ρ : V × I → I are the projection maps, then the composite mapping is C r and ϕ ϕ−1 Fτ π W0 → W2 → V × I → V → S has a tangent that is an isomorphism at each point, and so by the inverse mapping theorem, it is a C r diffeomorphism onto an open submanifold. Let W1 be its image, and Θ the composite mapping. We now show that Θ : W0 → W1 is a Poincar´e map. Obviously PM1 is satisfied. For PM2, we identify U and V × I by means of ϕ to simplify notations. Then π : W2 → W1 is a diffeomorphism, and its inverse

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(π|W2 )−1 : W1 → W2 ⊂ W1 × R is a section corresponding to a smooth function σ : W1 → R. In fact, σ is defined implicitly by Fτ (w0 ) = (π ◦ Fτ (w0 ), ρ ◦ Fτ (w0 )) = (π ◦ Fτ (w0 ), σ ◦ πFτ (w0 )) or ρ ◦ Fτ (w0 ) = σ ◦ πFτ (w0 ). Now let δ : W0 → R be given by w0 7→ σ ◦ Ft (w0 ) which is C r . Then we have Fτ −δ(w0 ) (w0 ) = (F−δ(w0 ) ◦ Fτ ) = (π ◦ Fτ (w0 ), ρ ◦ Fτ (w0 ) − δ(w0 )) = (π ◦ Fτ (w0 ), 0) = Θ(w0 ). Finally, PM3 is obvious since (U, ϕ) is a flow box. (ii) The proof is burdensome because of the notational complexity in the definition of local conjugacy, so we will be satisfied to prove uniqueness under additional simplifying hypotheses that lead to global conjugacy (identified by italics). The general case will be left to the reader. We consider first the special case m = m0 . Choose a flow box chart (U, ϕ) at m, and assume S ∪ S 0 ⊂ U , and that S and S 0 intersect each orbit arc in U at most once, and that they intersect exactly the same sets of orbits. (These three conditions may always be obtained by shrinking S and S 0 .) Then let W2 = S, W20 = S 0 , and H : W2 → W20 the bijection given by the orbits in U . As in (i), this is seen to be a C r diffeomorphism, and H ◦ Θ = Θ0 ◦ H. Finally, suppose m 6= m0 . Then Fa (m) = m0 for some a ∈ ]0, τ [, and as DX is open there is a neighborhood U of m such that U × {a} ⊂ DX . Then Fa (U ∩ S) = S 00 is a local transversal section of X at m0 ∈ γ, and H = Fa effects a conjugacy between Θ and Θ00 = Fa ◦ Θ ◦ Fa−1 on S 00 . By the preceding paragraph, Θ00 and Θ0 are locally conjugate, but conjugacy is an equivalence relation. This completes the argument.  If γ is a closed orbit of X ∈ X(M ) and m ∈ γ, the behavior of nearby orbits is given by a Poincar´e map Θ on a local transversal section S at m. Clearly Tm Θ ∈ L(Tm S, Tm S) is a linear approximation to Θ at m. By uniqueness of Θ up to local conjugacy, Tm0 Θ0 is similar to Tm Θ, for any other Poincar´e map Θ0 on a local transversal section at m0 ∈ γ. Therefore, the spectrum of Tm Θ is independent of m ∈ γ and the particular section S at m. 4.3.23 Definition. If γ is a closed orbit of X ∈ X(M ), the characteristic multipliers of X at γ are the points in the spectrum of Tm Θ, for any Poincar´e map Θ at any m ∈ γ. Another linear approximation to the flow near γ is given by Tm Fτ ∈ L(Tm M, Tm M ) if m ∈ γ and τ is the period of γ. Note that Fτ∗ (X(m)) = X(m), so Tm Fτ always has an eigenvalue 1 corresponding to the eigenvector X(m). The (n − 1) remaining eigenvalues (if dim(M ) = n) are in fact the characteristic multipliers of X at γ. 4.3.24 Proposition. If γ is a closed orbit of X ∈ X(M ) of period τ and cγ is the set of characteristic multipliers of X at γ, then cγ ∪ {1} is the spectrum of Tm Fτ , for any m ∈ γ. Proof. We can work in a chart modeled on E and assume m = 0. Let V be the span of X(m) so E = Tm M ⊕ V . Write the flow Ft (x, y) = (Ft1 (x, y), Ft2 (x, y)). By definition, we have D1 Ft1 (m) = Tm Θ

and

D2 Fτ2 (m) · X(m) = X(m).

Thus, the matrix of Tm Fτ is of the form "

Tm Θ

# 0

A

1

,

where A = D1 Fτ2 (m). From this it follows that the spectrum of Tm Fτ is {1} ∪ cγ .



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If the characteristic exponents of an equilibrium point lie (strictly) in the left half-plane, we know from Liapunov’s theorem that the equilibrium is stable. For closed orbits we introduce stability by means of the following definition. 4.3.25 Definition. Let X be a vector field on a manifold M and γ a closed orbit of X. An orbit Ft (m0 ) is said to wind toward γ if m0 is + complete and for any local transversal section S to X at m ∈ γ there is a t(0) such that Ft(0) (m0 ) ∈ S and successive applications of the Poincar´e map yield a sequence of points that converges to m. If the closed orbit γ has a neighborhood U such that for any m0 ∈ U , the orbit through m0 winds towards γ, then γ is called asymptotically stable. In other words, orbits starting “close” to γ, “converge” to γ; see Figure 4.3.10.

m0

Figure 4.3.10. Stable periodic orbit

4.3.26 Proposition. If γ is an asymptotically stable periodic orbit of the vector field X and m0 ∈ U , the neighborhood given in Definition 4.3.25, then for any neighborhood V of γ, there exists t0 > 0 such that for all t ≥ t0 , Ft (m) ∈ V . Proof. Define mk = Θk (m0 ), where Θ is a Poincar´e map for a local transversal section at m to γ containing m0 . Let t(n) be the “return time” of n, that is, t(n) is defined by Ft(n) (n) ∈ S. If τ denotes the period of γ and τk = t(mk ), then since mk → m, it follows that τk → τ since t(n) is a smooth function of n by Theorem 4.3.22. Let M be an upper bound for the set { |τk | | k ∈ N }. By smoothness of the flow, Fs (mk ) → Fs (m) as k → ∞, uniformly in s ∈ [0, M ]. Now write for any t > 0, Ft (m0 ) = FT (t) (mk(t) ), for T (t) ∈ [0, M ] and observe that k(t) → ∞ as t → ∞. Therefore, if W is any neighborhood of FT (t) (m0 ) contained in V , since Ft (m0 ) = FT (t) (mk(t) ) converges to FT (t) (m) as t → ∞ it follows that there exists t0 > 0 such that for all t ≥ t0 , Ft (m0 ) ∈ W ⊂ V .  4.3.27 Theorem (Liapunov Stability Theorem for Closed Orbits). Let γ be a closed orbit of X ∈ X(M ) and let the characteristic multipliers of γ lie strictly inside the unit circle. Then γ is asymptotically stable. The proof relies on the following lemma. 4.3.28 Lemma. Let T : E → E be a bounded linear operator. Let σ(T ) lie strictly inside the unit circle. Then limn→∞ T n e = 0 for all e ∈ E. Proof. By Lemma 4.3.7 and compactness of σ(T ), there is a norm | | on E equivalent to the original norm on E such that |T | ≤ r < 1. Therefore |T n e| ≤ rn |e| → 0 as n → ∞. H 4.3.29 Lemma. Let f : S → S be a smooth map on a manifold S with f (s) = s for some s. Let the spectrum of Ts f lie strictly inside the unit circle. Then there is a neighborhood U of s such that if s0 ∈ U , f (s0 ) ∈ U and f n (s0 ) → s as n → ∞, where f n = f ◦ f ◦ · · · ◦ f (n times). Proof. We can assume that S is a Banach space E and that s = 0. As above, renorm E and find 0 < r < 1 such that |T | ≤ r, where T = Df (0). Let ε > 0 be such that r + ε < 1. Choose a neighborhood V of 0 such that for all x ∈ V |f (x) − T x| ≤ ε|x|

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which is possible since f is smooth. Therefore, |f (x)| ≤ |T x| + ε|x| ≤ (r + ε)|x|. Now, choose δ > 0 such that the ball U of radius δ at 0 lies in V . Then the above inequality implies |f n (x)| ≤ (r + ε)n |x| for all x ∈ U which shows that f n (x) ∈ U and f n (x) → 0. H Proof of Theorem 4.3.27. The previous lemma applied to f = Θ, a Poincar´e map in a transversal slice S to γ at m, implies that there is an open neighborhood V of m in S such that the orbit through every point of V winds toward γ. Thus, the orbit through every point of U = { Ft (m) | t ≥ 0 } ⊃ γ winds toward γ. U is a neighborhood of γ since by the straightening out theorem 4.1.14, each point of γ has a neighborhood contained in { Ft (Θ(U )) | t > −ε, ε > 0 }.  4.3.30 Definition. If X ∈ X(M ) and γ is a closed orbit of X, γ is called hyperbolic if none of the characteristic multipliers of X at γ has modulus 1. Hyperbolic closed orbits are isolated (see Abraham and Robbin [1967, Chapter 5]). The local qualitative behavior near an hyperbolic closed orbit, γ, may be visualized with the aid of the Poincar´e map, Θ : W0 ⊂ S → W1 ⊂ S, as shown in Figure 4.3.8. The qualitative behavior of this map, under iterations, determines the asymptotic behavior of the trajectories near γ. Let m ∈ γ be the base point of the section, and s ∈ S. Then In(γ), the inset of γ, intersects S in the inset of m under the iterations of Θ. That is, s ∈ In(γ) if the trajectory Ft (s) winds towards γ, and this is equivalent to saying that Θk (s) tends to m as k → +∞. The inset and outset of m ∈ S are classified by linear algebra, as there is an analogue of the linearization theorem for maps at hyperbolic critical points. The linearization theorem for maps says that there is a C 0 coordinate chart on S, in which the local representative of Θ is a linear map. Recall that in the hyperbolic case, the spectrum of this linear isomorphism avoids the unit circle. The eigenvalues inside the unit circle determine the generalized eigenspace of contraction that is, the inset of m ∈ S under the iterates of Θ. The eigenvalues outside the unit circle similarly determine the outset of m ∈ S. Although this argument provides only local C 0 submanifolds, the global stable manifold theorem improves this: the inset and outset of a fixed point of a diffeomorphism are smooth, injectively immersed submanifolds meeting transversally at m. Returning to closed orbits, the inset and outset of γ ⊂ M may be visualized by choosing a section Sm at every point m ∈ γ. The inset and outset of γ in M intersect each section Sm in submanifolds of Sm , meeting transversally at m ∈ γ. In fact, In(γ) is a cylinder over γ, that is, a bundle of injectively immersed disks. So, likewise, is Out(γ). And these two cylinders intersect transversally in γ, as shown in Figure 4.3.11. These bundles need not be trivial.

Figure 4.3.11. Insets and outsets for periodic orbits

Another argument is sometimes used to study the inset and outset of a closed orbit, in place of the Poincar´e section technique described before, and is originally due to Smale [1967]. The flow Ft leaves the closed orbit invariant. A special coordinate chart may be found in a neighborhood of γ. The neighborhood

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is a disk bundle over γ, and the flow Ft , is a bundle map. On each fiber, Ft is a linear map of the form Zt eRt , where Zt is a constant, and R is a linear map. Thus, if s = (m, x) is a point in the chart, the local representative of Ft is given by the expression Ft (m, x) = (mt , Zt eRt · x) called the Floquet normal form. This is the linearization theorem for closed orbits. A related result, the Floquet theorem, eliminates the dependence of Zt on t, by making a further (time-dependent) change of coordinates (see Hartman [1973, Chapter 4, Section 6], or Abraham and Robbin [1967]). Finally, linear algebra applied to the linear map R in the exponent of the Floquet normal form, establishes the C 0 structure of the inset and outset of γ. To get an overall picture of a dynamical system in which all critical elements (critical points and closed orbits) are hyperbolic, we try to draw or visualize the insets and outsets of each. Those with open insets are attractors, and their open insets are their basins. The domain is divided into basins by the separatrices, which includes the insets of all the nonattractive (saddle-type) critical elements (and possible other, more complicated limit sets, called chaotic attractors, not described here.) We conclude with an example of sufficient complexity, which has been at the center of dynamical system theory for over a century. 4.3.31 Example. The simple pendulum equation may be “simplified” by approximating sin x by two terms of its MacLaurin expansion. The resulting system is a model for a nonlinear spring with linear damping , k x˙ = v, v˙ = −cv − kx + x3 . 3 Adding a periodic forcing term, we have x˙ = v,

v˙ = −cv − kx +

k 4 x + F cos ωt. 3

This time-dependent system in the plane is transformed into an autonomous system in a solid ring by adding an angular variable proportional to the time, θ = ωt. Thus, x˙ = v,

x˙ = −cv − kx +

k 3 x + F cos θ, 3

θ˙ = ω

Although this was introduced by Baron Rayleigh to study the resonance of tuning forks, piano strings, and so on, in his classic 1877 book, Theory of Sound , this system is generally named the Duffing equation after Duffing who obtained the first important results in 1908 (see Stoker [1950] for additional information).  Depending on the values of the three parameters (c, k, F ) various phase portraits are obtained. One of these is shown in Figure 4.3.12, adapted from the experiments of Hayashi [1964]. There are three closed orbits: two attracting, one of saddle type. The inset of the saddle is a cylinder topologically, but the whole cylinder revolves around the saddle-type closed orbit. This cylinder is the separatrix between the two basins. For other parameter values the dynamics can be chaotic (see for example, Holmes [1979a, 1979b] and Ueda [1980]). For further information on dynamical systems, see, for example, Guckenheimer and Holmes [1983].

Exercises  4.3-1.  4.3-2.

Prove that the equation θ¨ + 2k θ˙ − q sin θ = 0 (q > 0, k > 0) has a saddle point at θ = 0, θ˙ = 0. Consider the differential equations r˙ = ar3 − br, θ˙ = 1 using polar coordinates in the plane.

(i) Determine those a, b for which this system has an attractive periodic orbit.

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Figure 4.3.12. Phase portrait for the nonlinear spring with linear damping and forcing.

(ii) Calculate the eigenvalues of this system at the origin for various a, b.  4.3-3.

Let X ∈ X(M ), ϕ : M → N be a diffeomorphism, and Y = ϕ∗ X. Show that

(i) m ∈ M is a critical point of X iff ϕ(m) is a critical point of Y and the characteristic exponents are the same for each; (ii) γ ⊂ M is a closed orbit of X iff ϕ(γ) is a closed orbit of Y and their characteristic multipliers are the same.  4.3-4. This exercise generalizes the notion of the linearization of a vector field at a critical or equilibrium point to the corresponding notion for sections of vector bundles. (i) Let E → M be a vector bundle and m ∈ M an element of the zero section. Show that Tm E is isomorphic to Tm M ⊕ Em in a natural, chart independent way. (ii) If ξ : M → E is a section of E, and ξ(m) = 0, define ξ 0 (m) : Tm M → Em to be the projection of Tm ξ to Em . Write out ξ 0 (m) relative to coordinates. (iii) Show that if X is a vector field, then X 0 (m) defined this way coincides with Definition 4.3.1.  4.3-5.

The energy for a symmetric heavy top is H(θ, pθ ) =

 2 p2ψ 1 2 2 2 p (b − cos θ) + p sin θ + M g` cos θ ψ θ J 2I sin2 θ

where I, J > 0, b, pψ , and M g` > 0 are constants. The dynamics of the top is described by the differential equations θ˙ = ∂H/∂pθ , p˙θ = −∂H/∂θ. (i) Show that θ = 0, pθ = 0 is a saddle point if 0 < pψ < 2(M g`I)1/2 (a slow top). (ii) Verify that cos θ = 1 − γ sech2 ((βγ)1/2 /2), where γ = 2 − b2 /β and β = 2M g`/I describe both the outset and inset of this saddle point. (This is called a homoclinic orbit.)

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(iii) Is θ = 0, pθ = 0 stable if pψ > (M g`I)1/2 ? Hint: Use the fact that H is constant along the trajectories.  4.3-6. Let A ∈ L(Rn , Rn ) and suppose that a < Re(λi ) < b, for all eigenvalues λi , i = 1, . . . , n of A. Show that Rn admits an inner product hh , ii with associated norm ||| · ||| such that a|||x|||2 ≤ hhAx, xii ≤ b|||x|||2 . Prove this by following the outline below. (i) If A is diagonalizable over C, then find a basis in Rn in which the matrix of A has either the entries on the diagonal, the real eigenvalues of A, or 2 × 2 blocks of the form " # aj −bj , for λj = aj + ibj , if bj 6= 0 bj aj Choose the inner product hh , ii on Rn such that the one and two-dimensional invariant subspaces of A defined by this block-matrix are mutually orthogonal; pick the standard R2 -basis in the associated two-dimensional spaces. (ii) If A is not diagonalizable, then pass to the real Jordan blocks:  λj 0 0 ···   1 λj 0 ···    0 1 λj · · ·   · · ·  0 ··· ···

form. There are two kinds of Jordan kj × kj  0 0  0 0   0 0 ,  · · · · · ·  1 λj

if λj ∈ R, or 

∆j

0

0

···

0

  I2    0   · · ·  0

∆j

0

···

0

I2

∆j

···

0

where

···

···

aj

−bj

bj

aj

I2

#

" ,

I2 =



 0   0   · · · 

···

" ∆j =

0

∆j 1

# 0

0

1

,

if λj = aj +ibj , aj , bj ∈ R, bj 6= 0. For the first kind of block, choose the basis e01 , . . . , e0k(j) of eigenvectors of the diagonal part, which we call D. Then, for ε > 0 small, put eεr = e0r /er−1 , r = 1, . . . , k(j) and define h , iε on the subspace span{eε1 , . . . , eεk(j) } ⊂ Rn to be the Euclidean inner product given by this basis. Compute the matrix of A in this basis and show that hAx, xiε Dx · x → , hx, xiε kxk2

as

ε → 0.

Conclude that for ε small the statement holds for the first kind of block. Do the same for the second kind of block.  4.3-7.

Let A ∈ L(Rn , Rn ). Show that the following are equivalent.

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(i) All eigenvalues of A have strictly negative real part (the origin is called a sink in this case). (ii) For any norm | · | on Rn , there exist constants k > 0 and ε > 0 such that for all t ≥ 0, |etA | ≤ ke−tε . (iii) There is a norm ||| · ||| on Rn and a constant δ > 0 such that for all t ≥ 0, |||etA ||| ≤ e−tδ . Hint: (ii) ⇒ (i) by using the real Jordan form: if every solution of x˙ = Ax tends to zero as t → +∞, then every eigenvalue of A has strictly negative real part. For (i) ⇒ (iii) use Exercise 4.3-5 and observe that if x(t) is a solution of x˙ = Ax, then we have hhx(t), Ax(t)ii d |||x(t)||| = , dt |||x(t)||| so that we get the following inequality: at ≤ log |||x(t)|||/ log |||x(0)||| ≤ bt, where a = min{ Re(λi ) | i = 1, . . . , n }, and b = max{ Re(λi ) | i = 1, . . . , n }. Then let −ε = b. Prove a similar theorem if all eigenvalues of A have strictly positive real part; the origin is then called a source.  4.3-8. Give a proof of Theorem 4.3.4 in the finite dimensional case without using the variation of constants formula (Exercise 4.1-5) and using Exercise 4.3-6. Hint: If A = X 0 (0) locally show hhX(x) − Ax, xii lim = 0. x→0 |||x|||2 Since hhAx, xii ≤ −ε|||x|||2 , ε = max Re{ λi | i = 1, . . . , n, λi eigenvalues of A }, there exists δ > 0 such that if |||x||| ≤ δ, then hhX(x), xii ≤ −C|||x|||2 , for some C > 0. Show that if x(t) is a solution curve in the closed δ-ball, t ∈ [0, T ], then d|||x(t)||| ≤ −C|||x(t)|||. dt Conclude |||x(t)||| ≤ δ for all t ∈ [0, T ] and thus by compactness of the δ-ball, x(t) exists for all t ≥ 0. Finally, show that |||x(t)||| ≤ e−tε |||x(0)|||.  4.3-9. An equilibrium point m of a vector field X ∈ X(M ) is called a sink , if there is a δ > 0 such that all points in the spectrum of X 0 (m) have real part < −δ. (i) Show that in a neighborhood of a sink there is no other equilibrium of X. (ii) If M = Rn and X is a linear vector field, Exercise 4.3-7 shows that provided limt→∞ m(t) = 0 for every integral curve m(t) of X, then the eigenvalues of X have all strictly negative real part. Show that this statement is false for general vector fields by finding an example of a non-linear vector field X on Rn whose integral curves tend to zero as t → ∞, but is such that X 0 (0) has at least one eigenvalue with zero real part. Hint : See Exercise 4.3-2.  4.3-10 (Hyperbolic Flows). An operator A ∈ L(Rn , Rn ) is called hyperbolic if no eigenvalue of A has zero real part. The linear flow x 7→ etA x is then called a hyperbolic flow . (i) Let A be hyperbolic. Show that there is a direct sum decomposition Rn = Es ⊕ Eu , A(Es ) ⊂ Es , A(Eu ) ⊂ Eu , such that the origin is a sink on Es and a source on Eu ; Es and Eu are called the stable and unstable subspaces of Rn . Show that the decomposition is unique. Hint: Es is the sum of all subspaces defined by the real Jordan form for which the real part of the eigenvalues is negative. For uniqueness, if Rn = E0s ⊕ E0u and v ∈ E0s , then v = x + y, x ∈ Es , y ∈ Eu with etA v → 0 as t → ∞, so that etA x → 0, etA y → 0 as t → ∞. But since the origin is a source on E0u , ketA yk ≥ etε kyk for some ε > 0 by the analogue of Exercise 4.3-7(ii).

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(ii) Show that A is hyperbolic iff for each x 6= 0, ketA xk → ∞ as t → ±∞. (iii) Conclude that hyperbolic flows have no periodic orbits.  4.3-11 (Gradient flows; continuation of Exercise 4.1-8). Let f : Rn → R be C 1 and let X = (∂f /∂x1 , . . . , ∂f /∂xn ). Show that at regular points of f , the integral curves of X cross the level surfaces of f orthogonally and that every singular point of f is an equilibrium of X. Show that isolated maxima of f are asymptotically stable. Hint: If x0 is the isolated maximum of f , then f (x0 ) − f (x) is a strict Liapunov function. Draw the level sets of f and the integral curves of X on the same diagram in R2 , when f is defined by f (x1 , x2 ) = (x1 − 1)2 + (x2 − 2)2 (x2 − 3)2 . √  4.3-12. Consider u ¨ + u˙ + u3 = 0. Show that solutions converge to zero like C/ t as t → ∞ by considering H(u, u) ˙ = (u + u) ˙ 2 + u3 + u4 .  4.3-13. fields:

Use the method of Liapunov functions to study the stability of the origin for the following vector

(i) X(x, y) = −(3y + x3 )(∂/∂x) + (2x − 5y 3 )(∂/∂y) (asymptotically stable); (ii) X(x, y) = −xy 4 (∂/∂x) + x6 y(∂/∂y) (stable; look for L of the form x4 + ay 6 ); (iii) X(x, y) = (xy − x3 + y)(∂/∂x) + (x4 − x2 y + x3 )(∂/∂y) (stable; look for L of the form ax4 + by 2 ); (iv) X(x, y) = (y + x7 )(∂/∂x) + (y 9 − x)(∂/∂y) (unstable); (v) X(x, y, z) = 3y(z + 1)(∂/∂x) + x(z + 1)(∂/∂y) + yz(∂/∂z) (stable); (vi) X(x, y, z) = (−x5 + 5x6 + 2y 3 + xz 2 + xyz)(∂/∂x) + (−y − 2z + 3x6 + 4yz + xz + xy 2 )(∂/∂y) + (2y − z − 2x8 − y 2 + xz 2 + xy 3 )(∂/∂z) (asymptotically stable; use L(x, y, z) = (1/2)x2 + 5(y 2 + z 2 )).  4.3-14.

Consider the following vector field on Rn+1 ; X(s, x) = (asN + f (s) + g(s, x), Ax + F (x) + h(s, x))

where s ∈ R, x ∈ Rn , f (0) = · · · = f (N ) (0) = 0, g(s, x), has all derivatives of order ≤ 2 zero at the origin, and F (x), h(s, x) vanish together with their first derivative at the origin. Assume the n × n matrix A has all eigenvalues distinct with strictly negative real part. Prove the following theorem of Liapunov: if N is even or N is odd and a > 0, then the origin is unstable; if N is odd and a < 0, the origin is asymptotically stable. Hint: for N even, use L(s, x) = s − akxk2 /2 and for N odd, L(s, x) = (s2 − akxk2 )/2; show that in both cases the sign of X[L] near the origin is given by the sign of a.  4.3-15. Let E be a Banach space and A : R → L(E, E) a continuous map. Let Ft,s denote the evolution operator of the time-dependent vector field X(t, x) = A(t)x on E. (i) Show that Ft,s ∈ GL(E). (ii) Show that kFt,s k ≤ e(t−s)α , where α = supλ∈[s,t] kA(λ)k. Conclude that the vector field X(t, x) is complete. Hint: Use Gronwall’s inequality and the time dependent version of Proposition 4.1.22. Next assume that A is periodic with period T , that is, A(t + T ) = A(t) for all t ∈ R.

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(iii) Show that Ft+T,s+T = Ft,s for any t, s ∈ R. Hint: Show that t 7→ Ft+T,s+T (x) satisfies the differential equation x˙ = A(t)x. (iv) Define the monodromy operator by M (t) = Ft+T,t . Show that if A is independent of t, then M (t) = eT A is also independent of t. Show that M (s) = Ft,s ◦ M (t) ◦ Fs,t for any continuous A : R → L(E, E). Conclude that all solutions of x˙ = A(t)x are of period T if and only if there is a t0 such that M (t0 ) = identity. Show that the eigenvalues of M (t) are independent of t. (v) (Floquet). Show that each real eigenvalue λ of M (t0 ) determines a solution, denoted c(t; λ, t0 ) of x˙ = A(t)x satisfying c(t + T ; λ, t0 ) = λc(t; λ, t0 ), and also each complex eigenvalue λ = a + ib, of M (t0 ), where a, b ∈ R, determines a pair of solutions denoted cr (t; λ, t0 ) and ci (t; λ, t0 ) satisfying " r # " #" # c (t + T ; λ, t0 ) a −b cr (t; λ, t0 ) = . ci (t + T ; λ, t0 ) b a ci (t; λ, t0 ) Hint: Let c(t; λ, t0 ) denote the solution of x˙ = A(t)x with the initial condition c(t0 ; λ, t0 ) = e, where e is an eigenvector of M (t0 ) corresponding to λ; if λ is complex, work on the complexification of E. Then show that c(t + T ; λ, t0 ) − λc(t; λ, t0 ) satisfies the same differential equation and its value at t0 is zero since c(t0 + T ; λ, t0 ) = M (t0 )c(t0 ; λ, t0 ) = λc(t0 ; λ, t0 ).) (vi) (Floquet). Show that there is a nontrivial periodic solution of period T of x˙ = A(t)x if and only if 1 is an eigenvalue of M (t0 ) for some t0 ∈ R. Hint: If c(t) is such a periodic solution, then c(t0 ) = c(t0 + T ) = M (t0 )c(t0 ). (vii) (Liapunov). Let P : R → GL(E) be a C 1 function which is periodic with period T . Show that the change of variable y = P (t)x transforms the equation x˙ = A(t)x into the equation y˙ = B(t)y, where B(t) = (P 0 (t) + P (t)A(t))P (t)−1 . If N (t) is the monodromy operator of y˙ = B(t)y, show that N (t) = P (t)M (t)P (t)−1 .  4.3-16. (i) (Liapunov). Let E be a finite dimensional complex vector space and let A : R → L(E, E) be a continuous function which is periodic with period T . Let M (s) be the monodromy operator of the equation x˙ = A(t)x and let B ∈ L(E, E) be such that M (s) = eT B (see Exercise 4.1-15). Define P (t) = etB Fs,t and put y(t) = P (t)x(t). Use (vii) in the previous exercise to show that y(t) satisfies y˙ = By. Prove that P (t) is a periodic C 1 -function with period T . Hint: Use (iii) of the previous exercise and eT B = M (s) = Fs+T,s . Thus, for complex finite dimensional vector spaces, the equation x˙ = A(t)x, where A(t + T ) = A(t), can be transformed via y(t) = P (t)x(t) into the constant coefficient linear equation y˙ = By. (ii) Since the general solution of y˙ = By is a linear combination of vectors exp(tλi )t`(i) ui , where λ1 , . . . , λm are the distinct eigenvalues of B, ui ∈ E, and 1 ≤ `(i) ≤ multiplicity of λi , conclude that the general solution of x˙ = A(t)x is a linear combination of vectors exp(tλi )t`(i) P (t)−1 ui where P (t + T ) = P (t). Show that the eigenvalues of M (s) = eT B are exp(T λi ). Show that Re(λi ) < 0 (> 0) for all i = 1, . . . , m if and only if all the solutions of x˙ = A(t)x converge to the origin as t → +∞ (−∞), that is, if and only if the origin is asymptotically stable (unstable).

4.4

Frobenius’ Theorem and Foliations

The main pillars supporting differential topology and calculus on manifolds are the implicit function theorem, the existence theorem for ordinary differential equations, and Frobenius’ theorem, which we discuss briefly here. First some definitions.

4.4 Frobenius’ Theorem and Foliations

285

4.4.1 Definition. Let M be a manifold and let E ⊂ T M be a subbundle of its tangent bundle; that is, E is a distribution (or a plane field ) on M . (i) We say E is involutive if for any two vector fields X and Y defined on open sets of M and which take values in E, [X, Y ] takes values in E as well. (ii) We say E is integrable if for any m ∈ M there is a (local ) submanifold N ⊂ M , called a (local ) integral manifold of E at m containing m, whose tangent bundle is exactly E restricted to N . The situation is shown in Figure 4.4.1.

M0 N

Figure 4.4.1. Local integrable manifolds

4.4.2 Examples. A. Any subbundle E of T M with one dimensional fibers is involutive; E is also integrable, which is seen in the following way. Using local bundle charts for T M at m ∈ M with the subbundle property for E, find in an open neighborhood of m, and a vector field that never vanishes and has values in E. Its local integral curves through m have as their tangent bundles E restricted to these curves. If the vector field can be found globally and has no zeros, then through any point of the manifold there is exactly one maximal integral curve of the vector field, and this integral curve never reduces to a point. B. Let f : M → N be a submersion and consider the bundle ker T f ⊂ T M . This bundle is involutive since for any X, Y ∈ X(M ) which take values in ker T f , we have T f ([X, Y ]) = 0 by Proposition 4.2.25. The bundle is integrable since for any m ∈ M the restriction of ker T f to the submanifold f −1 (f (m)) coincides with the tangent bundle of this submanifold (see §3.5). C. Let Tn be the n-dimensional torus, n ≥ 2. Let 1 ≤ k ≤ n and consider E = { (v1 , . . . , vn ) ∈ T Tn | vk+1 = · · · = vn = 0 }. This distribution is involutive and integrable; the integral manifold through (t1 , . . . , tn ) is Tk × (tk+1 , . . . , tn ). D. E = T M is involutive and integrable; the integral submanifold through any point is M itself. A less trivial example consists of the two-dimensional span E of two linearly independent vector fields on R3 that commute, such as ∂ ∂ ∂ X=x +z ; Y =y ∂x ∂z ∂y

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on an open set (such as where x > 0, y > 0, z > 0) where X and Y are linearly independent. It is now easy to check that this distribution is involutive (see Example F below for the general statement). E. An example of a noninvolutive distribution is easy to find. For instance, one just needs to let E be the span of two linearly independent vector fields X and Y (so each fiber of E is two dimensional) and yet [X, Y ] is not in this span. For instance, let X=y

∂ ∂ − ; ∂x ∂y

Y =z

∂ ∂ −x ∂x ∂z

on a suitable set, such as the set where z > 0 on which these two vector fields are linearly independent, so define a distribution (again see Example F below for the general statement). One computes that [X, Y ] = −xy

∂ ∂z

which is not in E. Thus, this distribution is not integrable. A more elaborate, but also more interesting example is the following. Let M = SO(3), the rotation group (see Exercise 3.5-19). The tangent space at I = identity consists of the 3 × 3 skew symmetric matrices. Let     0 0 −q          for some p, q ∈ R 0 0 p EI = A ∈ TI SO(3) A =          q −p 0 a two dimensional subspace. For Q ∈ SO(3), let EQ = { B ∈ TQ SO(3) | Q−1 B ∈ EI }. S Then E = { EQ | Q ∈ SO(3) } is a distribution but is not involutive. In fact, one computes that the two vector fields with p = 1, q = 0 and p = 0, q = 1 have a bracket that does not lie in E. Further insight into this example is gained after one studies Lie groups (a supplementary chapter). F. Let E be a distribution on M . Suppose that a collection E of smooth sections of E spans E in the sense that for each section X of E there are vector fields X1 , . . . , Xk in E and smooth functions a1 , . . . , ak on M such that X = ai Xi . Suppose E is closed under bracketing; that is, if X and Y have values in E, so does [X, Y ]. We claim that E is involutive. To prove this assertion, let X and Y be sections of E and write X = ai Xi and Y = bj Yj , where ai and bj are smooth functions on M and Xi and Yj belong to E. We calculate [X, Y ] = B j Yj − Ai Xi + ai bj [Xi , Yj ], where B j = ai Xi [bj ]

and

Ai = bj Yj [ai ].

Thus [X, Y ] is a section of E, so E is involutive.



Frobenius’ theorem asserts that the two conditions in Definition 4.4.1 are equivalent. 4.4.3 Theorem (The Local Frobenius Theorem). integrable.

A subbundle E of T M is involutive if and only if it is

Proof. Suppose E is integrable. Let X and Y be sections of E and let N be a local integral manifold through m ∈ M . At points of N, X and Y are tangent to N , so define restricted vector fields X|N , Y |N on N . By Proposition 4.2.25 (on ϕ-relatedness of brackets) applied to the inclusion map, we have [X|N , Y |N ] = [X, Y ]|N .

4.4 Frobenius’ Theorem and Foliations

287

Since N is a manifold, [X|N , Y |N ] is a vector field on N , so [X, Y ] is tangent to N and hence in E. Conversely, suppose that E is involutive. By choosing a vector bundle chart, one is reduced to this local situation: E is a model space for the fibers of E, F is a complementary space, and U × V ⊂ E × F is an open neighborhood of (0, 0), so U × V is a local model for M . We have a map f : U × V → L(E, F) such that the fiber of E over (x, y) is E(x,y) = { (u, f (x, y) · u) | u ∈ E } ⊂ E × F, and we can assume we are working near (0, 0) and f (0, 0) = 0. Let us express involutivity of the distribution E in terms of f . For fixed u ∈ E, let Xu (x, y) = (u, f (x, y) · u). Using the local formula for the Lie bracket (see formula (4.2.9)) one finds that [Xu1 , Xu2 ](x, y) = (0, Df (x, y) · (u1 , f (x, y) · u1 ) · u2 −Df (x, y) · (u2 , f (x, y) · u2 ) · u1 ).

(4.4.1)

By the involution assumption, this lies in E(x,y) . Since the first component vanishes, the local description of E(x,y) above shows that the second must as well; that is, we get the following identity: Df (x, y) · (u1 , f (x, y) · u1 ) · u2 = Df (x, y) · (u2 , f (x, y) · u2 ) · u1 .

(4.4.2)

Consider the time-dependent vector fields Xt (x, y) = (0, f (tx, y) · x)

and Xt,u (x, y) = (u, f (tx, y) · tu),

so that by the local formula for the Jacobi–Lie bracket, [Xt , Xt,u ](x, y) = (0, tD2 f (tx, y) · (f (tx, y) · x) · u −tDf (tx, y) · (u, f (tx, y) · u) · x − f (tx, y) · u) = (0, −tD1 f (tx, y) · x · u − f (tx, y) · u),

(4.4.3)

where the last equality follows from (4.4.2). But ∂Xt,u /∂t equals the negative of the right hand side of equation (4.4.3), that is, ∂Xt,u = 0, [Xt , Xt,u ] + ∂t which by Theorem 4.2.31 is equivalent to d ∗ F Xt,u = 0, (4.4.4) dt t where Ft = Ft,0 and Ft,s is the evolution operator of the time dependent vector field Xt . Since Xt (0, 0) = 0, it follows that Ft is defined for 0 ≤ t ≤ 1 by Corollary 4.1.25. Since F0 is the identity, relation (4.4.4) implies that Ft∗ Xt,u = X0,u ,

i.e.,

T F1 ◦ X0,u = X1,u ◦ F1 .

(4.4.5)

Let N = F1 (E × {0}), a submanifold of E × F, the model space of M . If (x, y) = F1 (e, 0), the tangent space at (x, y) to N equals T(x,y) N = { T(e,0) F1 (u, 0)) | u ∈ E } = { T(e,0) F1 (X0,u (e, 0)) | u ∈ E } = { X1,u (F1 (e, 0)) | u ∈ E } (by (4.4.5)) = { (u, f (x, y) · u) | u ∈ E } = E(x,y) , as required.

(4.4.6) 

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The method of using the time-one map of a time-dependent flow to provide the appropriate coordinate change is useful in a number of situations and is called the method of Lie transforms. An abstract version of this method is given later in Example 6.4.7; we shall use this method again in Chapter 7 to prove the Poincar´e lemma and in Chapter 9 to prove the Darboux theorem. Note: The method of Lie transforms is also used in singularity theory and bifurcation theory (see Golubitsky and Schaeffer [1985]). For a proof of the Morse lemma using this method, see Proposition 6.5.8, which is based on Palais [1969] and Golubitsky and Marsden [1983]. For a proof of the Frobenius theorem from the implicit function theorem using manifolds of maps in the spirit of Supplement 4.1C, see Penot [1970]. See Exercise 4.4-7 for another proof of the Frobenius theorem in finite dimensions. Foliations. The Frobenius theorem is intimately connected to the global concept of foliations. Roughly speaking, the integral manifolds N can be glued together to form a “nicely stacked” family of submanifolds filling out M (see Figure 4.4.1 or Example 4.4.2A). 4.4.4 Definition. Let M be a manifold and Φ = {£α }α∈A a partition of M into disjoint connected sets called leaves. The partition Φ is called a foliation if each point of M has a chart (U, ϕ), ϕ : U → U 0 × V 0 ⊂ E ⊕ F such that for each £α the connected components (U ∩ £α )β of U ∩ £α are given by ϕ((U ∩ £α )β ) = U 0 × {cβα }, where cβα ∈ F are constants f or each α ∈ A and β. Such charts are called foliated (or distinguished ) by Φ. The dimension (respectively, codimension) of the foliation Φ is the dimension of E (resp., F ). See Figure 4.4.2.

U

y x

Figure 4.4.2. Chart for a foliation

Note that each leaf £α is a connected immersed submanifold. In general, this immersion is not an embedding; that is, the induced topology on £α from M does not necessarily coincide with the topology of £α (the leaf £α may accumulate on itself, for example). A differentiable structure on £α is induced by the foliated charts in the following manner. If (U, ϕ), ϕ : U → U 0 × V 0 ⊂ E ⊕ F is a foliated chart on M , and χ : E ⊕ F → F is the canonical projection, then χ ◦ ϕ restricted to (U ∩ £α )β defines a chart on £α . 4.4.5 Examples. A. The trivial foliation of a connected manifold M has only one leaf, M itself. It has codimension zero. If M is finite dimensional, both M and the leaf have the same dimension. Conversely, on a finite-dimensional connected manifold M , a foliation of dimension equal to dim(M ) is the trivial foliation. B. The discrete foliation of a manifold M is the only zero-dimensional foliation; its leaves are all points of M . If M is finite dimensional, the dimension of M is the codimension of this foliation. C. A vector field X that never vanishes on M determines a foliation; its leaves are the maximal integral curves of the vector field X. The fact that this is a foliation is the straightening out theorem (see §4.1).

4.4 Frobenius’ Theorem and Foliations

289

D. Let f : M → N be a submersion. It defines a foliation on M (of codimension equal to dim(N ) if dim(N ) is finite) by the collection of all connected components of f −1 (n) when n varies throughout N . The fact that this is a foliation is given by Theorem 3.5.4. In particular, we see that E ⊕ F is foliated by the family {E × {f }}f ∈F . E. In the preceding example, let M = R3 , N = R, and f (x1 , x2 , x3 ) = ϕ(r2 ) exp(x3 ), where ϕ : R → R is a C ∞ function satisfying ϕ(0) = 0, ϕ(1) = 0, and ϕ0 (s) < 0 for s > 0, and where r2 = (x1 )2 + (x2 )2 . Since df (x1 , x2 , x3 ) = exp(x3 )[2ϕ0 (r2 )x1 dx1 + 2ϕ0 (r2 )x2 dx2 + ϕ(r2 )dx3 ] and ϕ(r2 ) is a strictly decreasing function of r2 , f is submersion, so its level sets define a codimension one foliation on R3 . Since the only zero of ϕ(r2 ) occurs for r = 1, f −1 (0) equals the cylinder { (x1 , x2 , x3 ) | (x1 )2 + (x2 )2 = 1 }. Since ϕ(r2 ) is a positive function for r ∈ [0, 1[, it follows that if c > 0, then f −1 (c) = { (x1 , x2 , log(c/ϕ(r2 )) | 0 ≤ (x1 )2 + (x2 )2 < 1 }, which is diffeomorphic to the open unit ball in (x1 , x2 )-space via the projection (x1 , x2 , log(c/ϕ(r2 )) 7→ (x1 , x2 ). Note that the leaves f −1 (c), c > 0, are asymptotically tangent to the cylinder f −1 (0). Finally, since ϕ(r2 ) < 0 if r > 1, for c < 0 the leaves given by f −1 (c) = { (x1 , x2 , log(c/ϕ(r2 )) | (x1 )2 + (x2 )2 > 1 } are diffeomorphic to the plane minus the closed unit disk, that is, they are diffeomorphic to cylinders. As before, note that the cylinders f −1 (c) are asymptotically tangent to f −1 (0); see Figure 4.4.3.

Figure 4.4.3. A change in the topology of level sets.

F. (The Reeb Foliation on the Solid Torus and the Klein Bottle; Reeb [1952]). We claim that in the previous example, the leaves are in some sense translation invariant. The cylinder f −1 (0) is invariant; if c 6= 0, invariance is in the sense f −1 (c) + (0, 0, log t) = f −1 (tc), for any t > 0. Consider the part of the foliation within the solid cylinder (x1 )2 + (x2 )2 ≤ 1 and form the solid torus from this cylinder: identify (a1 , a2 , 0) with (b1 , b2 , 1) iff ai = bi , i = 1, 2. The foliation of the solid torus so obtained is called the orientable Reeb foliation. Out of the cylinder one can form the Klein bottle (see Figure 1.4.2) by considering the equivalence relation which identifies (a1 , a2 , 0) with (b1 , b2 , 1) iff a1 = b1 , a2 = −b2 . In this way one obtains the nonorientable Reeb foliation. (This terminology regarding orientability will be explained in §7.5.) G. (The Reeb Foliation on S 3 ; Reeb [1952]). Two orientable Reeb foliations on the solid torus determine a foliation on S 3 in the following way. The sphere S 3 is the union of two solid tori which are identified along their common boundary, the torus T2 , by the diffeomorphism taking meridians of one to parallels of the other and vice-versa. This foliation is called the Reeb foliation of S 3 ; it has one leaf diffeomorphic

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to the torus T2 and all its other leaves are diffeomorphic to R2 and accumulate on the torus. Below we describe, pictorially, the decomposition of S 3 in two solid tori. Remove the north pole, (1, 0, 0, 0), of S 3 and stereographically project the rest of S 3 onto R3 . In the plane (x2 , x4 ) draw two equal circles centered on the x2 -axis at the points a and b, where −a = b. Rotating about the x4 -axis yields the solid torus in R3 . Now draw all of the circles in the (x2 , x4 ) plane minus the two discs, centered on the x4 -axis and passing through a and b. Each such circle yields two connected arcs joining the two discs. In addition, consider the two portions of the x2 -axis: the line joining the two discs and the two rays going off from each disk separately. (See Figure 4.4.4.) Now rotate this figure about the x4 -axis. All arcs joining the disks generate smooth surfaces diffeomorphic to R2 and each such surface meets the solid torus along a parallel. Only the two rays emanating from the disks generate a surface diffeomorphic to the cylinder. Now add the north pole back to S 3 and pull back the whole structure via stereographic projection from R3 to S 3 : the cylinder becomes a torus and all surfaces diffeomorphic to R2 intersect this torus in meridians. Thus S 3 is the union of two solid tori glued along their common boundary by identifying parallels of one with meridians of the other and vice-versa. 

a

b

Figure 4.4.4. Construction for the Reeb foliation.

4.4.6 Proposition.

Let M be a manifold and Φ = {£α }α∈A be a foliation on M . The set T (M, Φ) =

[

[

Tm £α

α∈A m∈£α

is a subbundle of T M called the tangent bundle to the foliation. The quotient bundle, denoted ν(Φ) = T M/T (M, Φ), is called the normal bundle to the foliation Φ. Elements of T (M, Φ) are called vectors tangent to the foliation Φ. Proof. Let (U, ϕ), ϕ : U → U 0 × V 0 → E ⊕ F be a foliated chart. Since Tu ϕ(Tu £α ) = E × {0} for every u ∈ U ∩ £α , we have T ϕ(T U ∩ T (M, Φ)) = (U 0 × V 0 ) × (E × {0}). Thus, the standard tangent bundle charts induced by foliated charts of M have the subbundle property and naturally induce vector bundle charts by mapping vm ∈ Tm (M, Φ) to (ϕ(m), Tm ϕ(vm )) ∈ (U 0 × V 0 ) × (E × {0}). 

4.4 Frobenius’ Theorem and Foliations

4.4.7 Theorem (The Global Frobenius Theorem). equivalent:

291

Let E be a subbundle of T M . The following are

(i) There exists a foliation Φ on M such that E = T (M, Φ). (ii) E is integrable. (iii) E is involutive. Proof. The equivalence of (ii) and (iii) was proved in Theorem 4.4.3. Let (i) hold. Working with a foliated chart, E is integrable by Proposition 4.4.6, the integral submanifolds being the leaves of Φ. Thus (ii) holds. Finally, we need to show that (ii) implies (i). Consider on M the family of (local) integral manifolds of E, each equipped with its own submanifold topology. It is straightforward to verify that the family of finite intersections of open subsets of these local integral submanifolds defines a topology on M , finer in general than the original one. Let {£α }α∈A be its connected components. Then, denoting by (£α ∩ U )β the connected components of £α ∩ U in U , we have by definition E|(U ∩ £α )β = (E restricted to (U ∩ £α )β ) equals T ((U ∩ £α )β ). Let (τ −1 (U ), ψ), U ⊂ M be a vector bundle chart of T M with the subbundle property for E, and let ϕ : U → U1 be the induced chart on the base. This means, shrinking U if necessary, that ϕ : U → U 0 × V 0 ⊂ E ⊕ F and ψ(τ −1 (U ) ∩ E) = (U 0 × V 0 ) × (E × {0}). Thus, ϕ((U ∩ £α )β ) = U 0 × {cβα } and so M is foliated by the family Φ = {£α }α∈A . Because E|(U ∩ £α )β = T ((U ∩ £α )β ), we also have T (M, Φ) = E.  There is an important global topological condition that integrable subbundles must satisfy that was discovered by Bott [1970]. The result, called the Bott Vanishing Theorem, can be found, along with related results, by readers with background in algebraic topology, in Lawson [1977]. The leaves of a foliation are characterized by the following property. 4.4.8 Proposition. Let Φ be a foliation on M . Then x and y are in the same leaf if and only if x and y lie on the same integral curve of a vector field X defined on an open set in M and which is tangent to the foliation Φ. Proof. Let X be a vector field on M with values in T (M, Φ) and assume that x and y lie on the same integral curve of X. Let £ denote the leaf of Φ containing x. Since X is tangent to the foliation, X|£ is a vector field on £ and thus any integral curve of X starting in £ stays in £. Since y is on such an integral curve, it follows that y ∈ £. Conversely, let x, y ∈ £ and let c(t) be a smooth non-intersecting curve in £ such that c(0) = x, c(1) = y, c0 (t) 6= 0. (This can always be done on a connected manifold by showing that the set of points that can be so joined is open and closed.) Thus c : [0, 1] → £ is an immersion, and hence by compactness of [0, 1], c is an embedding. Using Definition 4.4.4, there is a neighborhood of the curve c in M which is diffeomorphic to a neighborhood of [0, 1] × {0} × {0} in R × F × G for Banach spaces F and G such that the leaves of the foliation have the local representation R × F × {w}, for fixed w ∈ G, and the image of the curve c has the local representation [0, 1] × {0} × {0}. Thus we can find a vector field X which is defined by c0 (t) along c and extends off c to be constant in this local representation.  Let R denote the following equivalence relation in a manifold M with a given foliation Φ : xRy if x, y belong to the same leaf of Φ. The previous proposition shows that R is an open equivalence relation. It is of interest to know whether M/R is a manifold. Foliations for which R is a regular equivalence relation are called regular foliations. (See §3.5 for a discussion of regular equivalence relations.) The following is a useful criterion.

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4.4.9 Proposition. Let Φ be a foliation on a manifold M and R the equivalence relation in M determined by Φ. R is regular iff for every m ∈ M there exists a local submanifold Σm of M such that Σm intersects every leaf in at most one point (or nowhere) and Tm Σm ⊕ Tm (M, Φ) = Tm M . (Sometimes Σm is called a slice or a local cross-section for the foliation.) Proof. Assume that R is regular and let π : M → M/R be the canonical projection. For Σm choose the submanifold using the following construction. Since π is a submersion, in appropriate charts (U, ϕ), (V, ψ), where ϕ : U → U 0 × V 0 and ψ : V → V 0 , the local representative of π, πϕψ : U 0 × V 0 → V 0 , is the projection onto the second factor, and every leaf π −1 (v) ⊂ U , v ∈ V , is represented in these charts as U 0 × {v 0 } where v 0 = ψ(v). Thus if Σm = ψ −1 ({0} × V 0 ), we see that Σm satisfies the two required conditions. Conversely, assume that each point m ∈ M admits a slice Σm . Working with a foliated chart, we are reduced to the following situation: let U, V be open balls centered at the origin in Banach spaces E and F , respectively, let Σ be a submanifold of U × V , (0, 0) ∈ Σ, such that T(0,0) Σ = F , and Σ ∩ (U × {v}) is at most one point for all v ∈ V . If p2 : E ⊕ F → F is the second projection, since p2 |Σ has tangent map at (0, 0) equal to the identity, it follows that for V small enough, p2 |Σ : Σ → V is a diffeomorphism. Shrinking Σ and V if necessary we can assume that Σ ∩ (U × {v}) is exactly one point. Let q : V → Σ be the inverse of p2 and define the smooth map s : U × V → Σ by s(u, v) = q(v). Then Σ ∩ (U × {v}) = {q(v)}, thus showing that Σ is a slice in the sense of Lemma 3.5.26. Pulling everything back to M by the foliated chart, the prior argument shows that for each point m ∈ M there is an open neighborhood U , a submanifold Σm of U , and a smooth map s : U → Σm such that £u ∩ Σm = {s(u)}, where £u is the leaf containing u ∈ U . By the argument following Lemma 3.5.26, the equivalence relation R is locally regular, that is, RU = R ∩ (U × U ) is regular. If U 0 = π −1 (π(U )) where π : M → M/R is the projection, the argument at the end of Step 1 in the proof of Theorem 3.5.25 shows that R is regular. Thus, all that remains to be proved is that T U can be chosen to equal U 0 . But this is clear by defining s0 : U 0 → Σm by s0 (u0 ) = s(u), where u ∈ U £u0 , £u0 being the leaf containing u0 ; smoothness of s0 follows from smoothness of s by composing it locally with the flow of a vector field given by Proposition 4.4.8.  To get a feeling for the foregoing condition we will study the linear flow on the torus. 4.4.10 Example. On the two-torus T2 consider the global flow F : R×T2 → T2 defined by F (t, (s1 , s2 )) = (s1 e2πit , s2 e2πiαt ) for a fixed number α ∈ [0, 1[. By Example 4.4.5C this defines a foliation on T2 . If α ∈ Q, notice that every integral curve is closed and that all integral curves have the same period. The condition of the previous theorem is easily verified and we conclude that in this case the equivalence relation R is regular; T2 /R = S 1 . If α is irrational, however, the situation is completely different. Let ϕ(t) = (e2πit , e2πiαt ) denote the integral curve through (1, 1). The following argument shows that cl (ϕ(R)) = T2 ; that is, ϕ(R) is dense in T2 . Let p = (e2πix , e2πiy ) ∈ T2 ; then for all m ∈ Z, ϕ(x + m) − p = (0, e2πiαx (e2πimα − e2πiz )) where y = αx + z. It suffices to show that C = { e2πimα ∈ S 1 | m ∈ Z } is dense in S 1 because then there is a sequence mk ∈ Z such that exp(2πimk α) converges to e2πiz . Hence, ϕ(x + mk ) converges to p. If for each k ∈ Z+ we divide S 1 into k arcs of length 2π/k, then, because { e2πimα ∈ S 1 | m = 1, 2, . . . , k + 1 } are distinct for some 1 ≤ nk < mk ≤ k + 1, exp(2πimk α) and exp(2πink α) belong to the same arc. Therefore, | exp(2πimk α) − exp(2πink α)| <

2π , k

which implies | exp(2πiqk α) − 1| < 2π/k, where qk = mk − nk . Because [ { e2πiαs ∈ S 1 | s ∈ [jqk , (j + 1)qk ] } = S 1 , j∈Z+

every arc of length less than 2π/k contains some exp(2πijqk ), which proves cl(C) = S 1 . Thus any submanifold Σm m ∈ T2 not coinciding with the integral curve through m will have to intersect ϕ(R) infinitely many

4.4 Frobenius’ Theorem and Foliations

293

times; the condition in the previous theorem is violated and so R is not regular. 

Remark. Novikov [1965] has shown that the Reeb foliation is in some sense typical. A foliation Φ on M is said to be transversally orientable if T M = T (M, Φ) ⊕ E, where E is an orientable subbundle of T M (see Exercise 7.5-14 for the definition). A foliation on a three dimensional manifold M is said to have a Reeb component if it has a compact leaf diffeomorphic to T2 or K and if the foliation within this torus or Klein bottle is diffeomorphic to the orientable or non-orientable Reeb foliation in the solid torus or the solid Klein bottle. Novikov has proved the following remarkable result. Let Φ be a transversally orientable C 2 codimension one foliation of a compact three dimensional manifold M . If π1 (M ) is finite, then Φ has a Reeb component, which may or may not be orientable. If π2 (M ) 6= 0 (with no hypotheses on π1 (M )) and Φ has no Reeb components, then all the leaves of Φ are compact with finite fundamental group. We refer the reader to Camacho and Neto [1985] for a proof of this result and to this reference and Lawson [1977] for a study of foliations in general. 

Stratifications. Even though foliations encompass “nice” partitions of a manifold into submanifolds, there are important situations when foliations are inappropriate because they are not regular or the leaves jump in dimension from point to point. Consider, for example, R2 as a union of concentric circles centered at the origin. As this example suggests, one would like to relax the condition that M/R be a manifold, provided that M/R turns out to be a union of manifolds that fit “nicely” together. Stratifications, another concept allowing us to “stack” manifolds, turn out to be the natural tool to describe the topology of orbit spaces of compact Lie group actions or non-compact Lie group actions admitting a slice (see, for instance, Bredon [1972], Burghelea, Albu, and Ratiu [1975], Fischer [1970], and Bourguigon [1975]). We shall limit ourselves to the definition in the finite-dimensional case and some simple remarks. 4.4.11 Definition. Let M be a locally compact topological space. A stratification of M is a partition of M into manifolds {Ma }a∈A called strata, satisfying the following conditions: S1. Ma are manifolds of constant dimension; they are submanifolds of M if M is itself a manifold. S2. The family {Maα } of connected components of all the Ma is a locally finite partition of M ; that is, for every m ∈ M , there exists an open neighborhood U of m in M intersecting only finitely many Mbβ . S3. If Maα ∩ cl(Mbβ ) 6= ∅ for (a, α) 6= (b, β), then Maα ⊂ Mbβ and dim(Ma ) < dim(Mb ). S4. cl(Ma )\Ma is a disjoint union of strata of dimension strictly less than dim(Ma ). From the definition it follows that if Maα ∩cl(Mbβ ) 6= ∅ and if m ∈ Maα ⊂ cl(Mbβ ) has an open neighborhood U in the topology of Maα such that U ⊂ Maα ∩ cl(Mbβ ), then necessarily Maα ⊂ Mbβ and thus dim(Ma ) < dim(Mb ). To see this, it is enough to note that the given hypothesis makes Maα ∩ cl(Mbβ ) open in Maα . Since it is also closed (by definition of the relative topology) and Maα is connected, it must equal Maα itself, whence Maα ⊂ cl(Mbβ ) and by S3, Maα ⊂ Mbβ and dim(Ma ) < dim(Mb ). For nonregular equivalence relations R, M/R is often a stratified space. The intuitive idea is that it is often possible to group together equivalence classes of the same dimension, and this grouping is parametrized by a manifold, which will be a stratum in M/R. A simple example is R2 partitioned by circles (the equivalence classes for R). The circles of positive radius are parametrized by the interval ]0, ∞[. Thus M/R is the stratified set [0, ∞[ consisting of the two strata {0} and ]0, ∞[.

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Exercises  4.4-1. (i) Give an explicit example of a three-dimensional nonintegrable distribution on the whole of R4 . (ii) Find an explicit example of a two dimensional integrable distribution on SO(3).  4.4-2. Let M be an n-manifold such that T M = E1 ⊕ · · · ⊕ Ep , where Ei , i = 1, . . . , p is an involutive subbundle of T M . Show that there are subspaces Ei ⊂ Rn , i = 1, . . . , p such that Rn = E1 ⊕ . . . ⊕ Ep and local charts ϕ : U ⊂ M → V ⊂ Rn , such that T ϕ maps each fiber of Ei onto Ei .  4.4-3. In R4 consider the family of surfaces given by x2 + y 2 + z 2 − t2 = const. Show that these surfaces define a stratification. What part of R4 should be thrown out to obtain a regular foliation?  4.4-4. Let f : M → N be a C ∞ map and Φ a foliation on N . The map f is said to be transversal to Φ, denoted f t Φ, if for every m ∈ M , Tm f (Tm M ) + Tf (m) (N, Φ) = Tf (m) N

and (Tm f )−1 (Tf (m) (N, Φ))

splits in Tm M . Show that if {£α }α∈A are the leaves of Φ, the connected components of f −1 (£α ) are leaves of a foliation (denoted by f ∗ (Φ)) on M , and if Φ has finite codimension in N , then so does the foliation f ∗ (Φ) on M and the two codimensions coincide.  4.4-5 (Bourbaki [1971]). Let M be a manifold and denote by M 0 the manifold with underlying set M but with a different differentiable structure. Show that the collection of connected components of M 0 defines a foliation of M iff for every m ∈ M , there exists an open set U in M , m ∈ U , a manifold N , and a submersion ρ : U → N such that the submanifold ρ−1 (n) of U is open in M 0 for all n ∈ N . Hint: For the “if” part use Lemma 3.3.5 and for the “only if” part use Exercise 3.2-6 to define a manifold structure on the leaves; the charts of the second structure are (U ∩ £α )β → U 0 .  4.4-6. On the manifold SO(3), consider the partition £A = { QA | Q is an arbitrary rotation about the z-axis in R3 }, A ∈ SO(3). Show that Φ = { £A | A ∈ SO(3) } is a regular foliation and that the quotient manifold SO(3)/R is diffeomorphic to S 2 .  4.4-7 (Hirsch and Weinstein). Give another proof of the Frobenius theorem as follows: Step 1. Prove it for the Abelian case in which all sections of E satisfy [X, Y ] = 0 by choosing a local basis X1 , . . . , Xk of sections and successively flowing out by the commuting flows of X1 , . . . , Xk . Step 2 Given a k-dimensional plane field, locally write it as a “graph” over Rk . Choose k commuting vector fields on Rk and lift them to the plane field. If E is involutive, the bracket of two of them lies in E and, moreover, since the bracket “pushes down” to Rk (by “relatedness”), it is zero. (This is actually demonstrated in formula (4.4.1).) Now use Step 1.

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5 An Introduction to Lie Groups

This version: January 31, 2007

Lie groups are groups in the sense of algebra and which are also manifolds such that the group operations are smooth maps. Lie groups, which were developed in the fundametal treatise of Lie [1890] are very important in a wide variety of applications, such as in understanding the emergence of patterns in biological systems, materials and other systems, the relation between symmetry and conserved quantities in mechanics and in gauge theory in Physics. This chapter concentrates on elementary aspects of the theory and discusses some of the basic groups, such as the rotation and Euclidean groups. There are many topics in the general Lie groups that we do not deal with here and more advanced treatises on the subject should be consulted, such as Warner [1983] and Knapp [2002]. For applications of Lie groups, we refer to one of the many books on the subject such as Sattinger and Weaver [1986]; Naber [2000]; Arnold [1989]; Libermann and Marle [1987]; Marsden and Ratiu [1999].

5.1

Basic Definitions and Properties

5.1.1 Definition. A Lie group is a (Banach) manifold G that has a group structure consistent with its manifold structure in the sense that group multiplication µ : G × G → G, is a C

∞

(g, h) 7→ gh,

map.

The maps Lg : G → G, h 7→ gh, and Rh : G → G, g → 7 gh, are called the left and right translation maps. Note that Lg1 ◦ Lg2 = Lg1 g2 and Rh1 ◦ Rh2 = Rh2 h1 . If e ∈ G denotes the identity element, then Le = Id = Re , and so (Lg )−1 = Lg−1

and

(Rh )−1 = Rh−1 .

Thus, Lg and Rh are diffeomorphisms for each g and h. Notice that Lg ◦ Rh = Rh ◦ Lg ,

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that is, left and right translations commute. By the chain rule, Tgh Lg−1 ◦ Th Lg = Th (Lg−1 ◦ Lg ) = Id . Thus, Th Lg is invertible. Likewise, Tg Rh is an isomorphism. We now show that the inversion map I : G → G; g 7→ g −1 is C ∞ . Indeed, consider solving µ(g, h) = e for h as a function of g. The partial derivative with respect to h is just Th Lg , which is an isomorphism. Thus, the solution g −1 is a smooth function of g by the implicit function theorem. Lie groups can be finite- or infinite-dimensional. For a first reading of this section, the reader may wish to assume that G is finite-dimensional.1

Examples (a)

Any Banach space E is an Abelian Lie group with group operations µ : E × E → E,

µ(x, y) = x + y,

and I : E → E,

I(x) = −x.

The identity is just the zero vector. We call such a Lie group a vector group.



(b) The group of linear isomorphisms of Rn to Rn is a Lie group of dimension n2 , called the general linear group and denoted by GL(n, R). It is a smooth manifold, since it is an open subset of the vector space L(Rn , Rn ) of all linear maps of Rn to Rn . Indeed, GL(n, R) is the inverse image of R\{0} under the continuous map A 7→ det A of L(Rn , Rn ) to R. For A, B ∈ GL(n, R), the group operation is composition, µ : GL(n, R) × GL(n, R) → GL(n, R) given by (A, B) 7→ A ◦ B, and the inversion map is I : GL(n, R) → GL(n, R) defined by I(A) = A−1 . Group multiplication is the restriction of the continuous bilinear map (A, B) ∈ L(Rn , Rn ) × L(Rn , Rn ) 7→ A ◦ B ∈ L(Rn , Rn ). Thus, µ is C ∞ , and so GL(n, R) is a Lie group. The group identity element e is the identity map on Rn . If we choose a basis in Rn , we can represent each A ∈ GL(n, R) by an invertible n × n matrix. The group operation is then matrix multiplication µ(A, B) = AB, and I(A) = A−1 is matrix inversion. The identity element e is the n × n identity matrix. The group operations are obviously smooth, since the formulas for the product and inverse of matrices are smooth (rational) functions of the matrix components.  (c) In the same way, one sees that for a Banach space E, the group GL(E, E) of invertible elements of L(E, E) is a Banach–Lie group. In Chapter 2 we proved that GL(E, E) is open in L(E, E). Further examples are given in the next section.  1 We caution that some interesting infinite-dimensional groups (such as groups of diffeomorphisms) are not literally Banach– Lie groups in the (naive) sense just given.

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297

Charts. Given any local chart on G, one can construct an entire atlas on the Lie group G by use of left (or right) translations. Suppose, for example, that (U, ϕ) is a chart about e ∈ G, and that ϕ : U → V . Define a chart (Ug , ϕg ) about g ∈ G by letting Ug = Lg (U ) = { Lg h | h ∈ U } and defining ϕg = ϕ ◦ Lg−1 : Ug → V, h 7→ ϕ(g −1 h). The set of charts {(Ug , ϕg )} forms an atlas, provided that one can show that the transition maps −1 : ϕg2 (Ug1 ∩ Ug2 ) → ϕg1 (Ug1 ∩ Ug2 ) ϕg1 ◦ ϕ−1 g2 = ϕ ◦ Lg −1 g2 ◦ ϕ 1

are diffeomorphisms (between open sets in a Banach space). But this follows from the smoothness of group multiplication and inversion. Invariant Vector Fields. A vector field X on G is called left invariant if for every g ∈ G we have L∗g X = X, which is equivalent to (Lg )∗ X = X. This is in turn equivalent to saying that (Th Lg )X(h) = X(gh) for every h ∈ G. We have the commutative diagram in Figure 5.1.1 and illustrate the geometry in Figure 5.1.2.

X

TG 6

T Lg -

G

-

TG 6 X G

Lg Figure 5.1.1. The commutative diagram for a left-invariant vector field.

h

gh

Figure 5.1.2. A left-invariant vector field.

Let XL (G) denote the set of left-invariant vector fields on G. If g ∈ G and X, Y ∈ XL (G), then L∗g [X, Y ] = [L∗g X, L∗g Y ] = [X, Y ], so [X, Y ] ∈ XL (G). Therefore, XL (G) is a Lie subalgebra of X(G), the set of all vector fields on G.

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For each ξ ∈ Te G, we define a vector field Xξ on G by letting Xξ (g) = Te Lg (ξ). Then Xξ (gh) = Te Lgh (ξ) = Te (Lg ◦ Lh )(ξ) = Th Lg (Te Lh (ξ)) = Th Lg (Xξ (h)), which shows that Xξ is left invariant. The linear maps ζ1 : XL (G) → Te G,

X 7→ X(e)

and ζ2 : Te G → XL (G),

ξ 7→ Xξ

satisfy ζ1 ◦ ζ2 = idTe G and ζ2 ◦ ζ1 = idXL (G) . Therefore, XL (G) and Te G are isomorphic as vector spaces. The Lie Algebra of a Lie Group.

Define the Lie bracket in Te G by [ξ, η] := [Xξ , Xη ](e),

where ξ, η ∈ Te G and where [Xξ , Xη ] is the Jacobi–Lie bracket of vector fields. This clearly makes Te G into a Lie algebra. Recall that Lie algebras were defined in Chapter 4 where we showed that the space of vector fields on a manifold is a Lie algebra under the Jacobi-Lie bracket. We say that this defines a bracket in Te G via left extension. Note that by construction, [Xξ , Xη ] = X[ξ,η] for all ξ, η ∈ Te G. 5.1.2 Definition. is denoted by g.

The vector space Te G with this Lie algebra structure is called the Lie algebra of G and

Defining the set XR (G) of right-invariant vector fields on G in the analogous way, we get a vector space isomorphism ξ 7→ Yξ , where Yξ (g) = (Te Rg )(ξ), between Te G = g and XR (G). In this way, each ξ ∈ g defines an element Yξ ∈ XR (G), and also an element Xξ ∈ XL (G). We will prove that a relation between Xξ and Yξ is given by I∗ Xξ = −Yξ , (5.1.1) where I : G → G is the inversion map: I(g) = g −1 . Since I is a diffeomorphism, (5.1.1) shows that I∗ : XL (G) → XR (G) is a vector space isomorphism. To prove (5.1.1) notice first that for u ∈ Tg G and v ∈ Th G, the derivative of the multiplication map has the expression T(g,h) µ(u, v) = Th Lg (v) + Tg Rh (u).

(5.1.2)

In addition, differentiating the map g 7→ µ(g, I(g)) = e gives T(g,g−1 ) µ(u, Tg I(u)) = 0 for all u ∈ Tg G. This and (5.1.2) yield Tg I(u) = −(Te Rg−1 ◦ Tg Lg−1 )(u),

(5.1.3)

for all u ∈ Tg G. Consequently, if ξ ∈ g, and g ∈ G, we have (I∗ Xξ )(g) = (T I ◦ Xξ ◦ I −1 )(g) = Tg−1 I(Xξ (g −1 )) = −(Te Rg ◦ Tg−1 Lg )(Xξ (g −1 )) = −Te Rg (ξ) = −Yξ (g)

(by (5.1.3)) (since Xξ (g

−1

) = Te Lg−1 (ξ))

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299

and (5.1.1) is proved. Hence for ξ, η ∈ g, −Y[ξ,η] = I∗ X[ξ,η] = I∗ [Xξ , Xη ] = [I∗ Xξ , I∗ Xη ] = [−Yξ , −Yη ] = [Yξ , Yη ], so that −[Yξ , Yη ](e) = Y[ξ,η] (e) = [ξ, η] = [Xξ , Xη ](e). Therefore, the Lie algebra bracket [ , ]R in g defined by right extension of elements in g, [ξ, η]R := [Yξ , Yη ](e), is the negative of the one defined by left extension, that is, [ξ, η]R := −[ξ, η].

Examples (a) For a vector group E, Te E ∼ = E and the left-invariant vector field defined by u ∈ Te E is the constant vector field Xu (v) = u for all v ∈ E. Therefore, the Lie algebra of a vector group E is E itself, with the trivial bracket [v, w] = 0 for all v, w ∈ E. We say that the Lie algebra is Abelian in this case.  (b) The Lie algebra of GL(n, R) is L(Rn , Rn ), also denoted by gl(n), the vector space of all linear transformations of Rn , with the commutator bracket [ξ, η] = ξη − ηξ. To see this, we recall that GL(n, R) is open in L(Rn , Rn ), and so the Lie algebra, as a vector space, is L(Rn , Rn ). To compute the bracket, note that for any ξ ∈ L(Rn , Rn ), Xξ : GL(n, R) → L(Rn , Rn ) given by A 7→ Aξ is a left-invariant vector field on GL(n, R) because for every B ∈ GL(n, R), the map LB : GL(n, R) → GL(n, R) defined by LB (A) = BA is a linear mapping, and hence Xξ (LB A) = BAξ = TA LB Xξ (A). Therefore, by the local formula [X, Y ](x) = DY (x) · X(x) − DX(x) · Y (x), we get [ξ, η] = [Xξ , Xη ](I) = DXη (I) · Xξ (I) − DXξ (I) · Xη (I). But Xη (A) = Aη is linear in A, so DXη (I) · B = Bη. Hence DXη (I) · Xξ (I) = ξη, and similarly DXξ (I) · Xη (I) = ηξ. Thus, L(Rn , Rn ) has the bracket [ξ, η] = ξη − ηξ.

(5.1.4) 

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(c) We can also establish (5.1.4) by a coordinate calculation. Choosing a basis in Rn , each A ∈ GL(n, R) is specified by its components Aij such that (Av)i = Aij v j (sum on j). Thus, a vector field X on GL(n, R) P has the form X(A) = i,j Cji (A)(∂/∂Aij ). It is checked to be left invariant, provided that there is a matrix (ξji ) such that for all A, X ∂ X(A) = Aik ξjk . ∂Aij i,j,k P If Y (A) = i,j,k Aik ηjk (∂/∂Aij ) is another left-invariant vector field, we have  X X i k ∂ l m ∂f (XY )[f ] = Ak ξj Am ηp ∂Alp ∂Aij X j m ∂f = Aik ξjk δil δm ηp + (second derivatives) ∂Alp X j ∂f = Aik ξjk ηm + (second derivatives), ∂Aij j . Therefore, the bracket is the left-invariant vector field [X, Y ] given where we have used ∂Asm /∂Akj = δsk δm by X ∂f j j [X, Y ][f ] = (XY − Y X)[f ] = Aik (ξjk ηm − ηjk ξm ) i . ∂Am

This shows that the vector field bracket is the usual commutator bracket of n × n matrices, as before.



One-Parameter Subgroups and the Exponential Map. If Xξ is the left-invariant vector field corresponding to ξ ∈ g, there is a unique integral curve γξ : R → G of Xξ starting at e, γξ (0) = e and γξ0 (t) = Xξ (γξ (t)). We claim that γξ (s + t) = γξ (s)γξ (t), which means that γξ (t) is a smooth one-parameter subgroup. Indeed, as functions of t, both sides equal γξ (s) at t = 0 and both satisfy the differential equation σ 0 (t) = Xξ (σ(t)) by left invariance of Xξ , so they are equal. Left invariance or γξ (t + s) = γξ (t)γξ (s) also shows that γξ (t) is defined for all t ∈ R. 5.1.3 Definition.

The exponential map exp : g → G is defined by exp(ξ) = γξ (1).

We claim that exp(sξ) = γξ (s). Indeed, for fixed s ∈ R, the curve t 7→ γξ (ts), which at t = 0 passes through e, satisfies the differential equation d γξ (ts) = sXξ (γξ (ts)) = Xsξ (γξ (ts)). dt Since γsξ (t) satisfies the same differential equation and passes through e at t = 0, it follows that γsξ (t) = γξ (ts). Putting t = 1 yields exp(sξ) = γξ (s). Hence the exponential mapping maps the line sξ in g onto the one-parameter subgroup γξ (s) of G, which is tangent to ξ at e. It follows from left invariance that the flow Ftξ of Xξ satisfies Ftξ (g) = gFtξ (e) = gγξ (t), so Ftξ (g) = g exp(tξ) = Rexp tξ g. Let γ(t) be a smooth one-parameter subgroup of G, so γ(0) = e in particular. We claim that γ = γξ , where ξ = γ 0 (0). Indeed, taking the derivative at s = 0 in the relation γ(t + s) = γ(t)γ(s) gives d dγ(t) = Lγ(t) γ(s) = Te Lγ(t) γ 0 (0) = Xξ (γ(t)), dt ds s=0

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301

so that γ = γξ , since both equal e at t = 0. In other words, all smooth one-parameter subgroups of G are of the form exp tξ for some ξ ∈ g. Since everything proved above for Xξ can be repeated for Yξ , it follows that the exponential map is the same for the left and right Lie algebras of a Lie group. From smoothness of the group operations and smoothness of the solutions of differential equations with respect to initial conditions, it follows that exp is a C ∞ map. Differentiating the identity exp(sξ) = γξ (s) with respect to s at s = 0 shows that T0 exp = idg . Therefore, by the inverse function theorem, exp is a local diffeomorphism from a neighborhood of zero in g onto a neighborhood of e in G. In other words, the exponential map defines a local chart for G at e; in finite dimensions, the coordinates associated to this chart are called the canonical coordinates of G. By left translation, this chart provides an atlas for G. (For typical infinite-dimensional groups like diffeomorphism groups, exp is not locally onto a neighborhood of the identity. It is also not true that the exponential map is a local diffeomorphism at any ξ 6= 0, even for finite-dimensional Lie groups.) It turns out that the exponential map characterizes not only the smooth one-parameter subgroups of G, but the continuous ones as well, as given in the next proposition (see for example, Varadarajan [1974] for the proof). 5.1.4 Proposition. Let γ : R → G be a continuous one-parameter subgroup of G. Then γ is automatically smooth, and hence γ(t) = exp tξ, for some ξ ∈ g.

Examples (a) Let G = V be a vector group, that is, V is a vector space and the group operation is vector addition. Then g = V and exp : V → V is the identity mapping.  (b) Let G = GL(n, R); so g = L(Rn , Rn ). For every A ∈ L(Rn , Rn ), the mapping γA : R → GL(n, R) defined by ∞ i X t i A t 7→ i! i=0 is a one-parameter subgroup, because γA (0) = I and 0 γA (t) =

∞ X ti−1 Ai = γA (t)A. (i − 1)! i=0

Therefore, the exponential mapping is given by exp : L(Rn , Rn ) → GL(n, Rn ),

A 7→ γA (1) =

∞ X Ai i=0

i!

.

As is customary, we will write eA =

∞ X Ai i=0

i!

.

We sometimes write expG : g → G when there is more than one group involved.



(c) Let G1 and G2 be Lie groups with Lie algebras g1 and g2 . Then G1 × G2 is a Lie group with Lie algebra g1 × g2 , and the exponential map is given by exp : g1 × g2 → G1 × G2 ,

(ξ1 , ξ2 ) 7→ (exp1 (ξ1 ), exp2 (ξ2 )).



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5. An Introduction to Lie Groups

Computing Brackets. steps:

Here is a computationally useful formula for the bracket. One follows these three

1. Calculate the inner automorphisms Ig : G → G, where Ig (h) = ghg −1 . 2. Differentiate Ig (h) with respect to h at h = e to produce the adjoint operators Adg : g → g;

Adg η = Te Ig · η.

Note that (see Figure 5.1.3) Adg η = Tg Rg−1 · Te Lg · η = Tg−1 Lg · Te Rg−1 · η. 3. Differentiate Adg η with respect to g at e in the direction ξ to get [ξ, η], that is, Te ϕη · ξ = [ξ, η],

(5.1.5)

where ϕη (g) = Adg η.

g e

Figure 5.1.3. The adjoint mapping is the linearization of conjugation.

5.1.5 Proposition. Proof.

Formula (5.1.5) is valid.

Denote by ϕt (g) = g exp tξ = Rexp tξ g the flow of Xξ . Then d −1 Tϕt (e) ϕt · Xη (ϕt (e)) [ξ, η] = [Xξ , Xη ](e) = dt t=0 d = Texp tξ Rexp(−tξ) Xη (exp tξ) dt t=0 d = Texp tξ Rexp(−tξ) Te Lexp tξ η dt t=0 d = Te (Lexp tξ ◦ Rexp(−tξ) )η dt t=0 d = Adexp tξ η , dt t=0

which is (5.1.5).



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303

Another way of expressing (5.1.5) is [ξ, η] =

d d , g(t)h(s)g(t)−1 dt ds s=0,t=0

(5.1.6)

where g(t) and h(s) are curves in G with g(0) = e, h(0) = e, and where g 0 (0) = ξ and h0 (0) = η. Example. and so

Consider the group GL(n, R). Formula (5.1.4) also follows from (5.1.5). Here, IA B = ABA−1 , AdA η = AηA−1 .

Differentiating this with respect to A at A = Identity in the direction ξ gives [ξ, η] = ξη − ηξ. Group Homomorphisms.



Some simple facts about Lie group homomorphisms will prove useful.

5.1.6 Proposition. Let G and H be Lie groups with Lie algebras g and h. Let f : G → H be a smooth homomorphism of Lie groups, that is, f (gh) = f (g)f (h), for all g, h ∈ G. Then Te f : g → h is a Lie algebra homomorphism, that is, (Te f )[ξ, η] = [Te f (ξ), Te f (η)], for all ξ, η ∈ g. In addition, f ◦ expG = expH ◦Te f. Proof. Since f is a group homomorphism, f ◦ Lg = Lf (g) ◦ f . Thus, T f ◦ T Lg = T Lf (g) ◦ T f , from which it follows that XTe f (ξ) (f (g)) = Tg f (Xξ (g)), that is, Xξ and XTe f (ξ) are f -related . It follows that the vector fields [Xξ , Xη ] and [XTe f (ξ) , XTe f (η) ] are also f -related for all ξ, η ∈ g. Hence Te f ([ξ, η]) = (T f ◦ [Xξ , Xη ])(e) = [XTe f (ξ) , XTe f (η) ](¯ e) = [Te f (ξ), Te f (η)].

(where e = eG ) (where e¯ = eH = f (e))

Thus, Te f is a Lie algebra homomorphism. Fixing ξ ∈ g, note that α : t 7→ f (expG (tξ)) and β : t 7→ expH (tTe f (ξ)) are one-parameter subgroups of H. Moreover, α0 (0) = Te f (ξ) = β 0 (0), and so α = β. In particular, f (expG (ξ)) = expH (Te f (ξ)), for all ξ ∈ g.  Example.

Proposition 5.1.6 applied to the determinant map gives the identity det(exp A) = exp(trace A)

for A ∈ GL(n, R).



5.1.7 Corollary. Assume that f1 , f2 : G → H are homomorphisms of Lie groups and that G is connected. If Te f1 = Te f2 , then f1 = f2 . This follows from Proposition 5.1.6, since a connected Lie group G is generated by a neighborhood of the identity element. This latter fact may be proved following these steps: 1. Show that any open subgroup of a Lie group is closed (since its complement is a union of group cosets, each of which is homeomorphic to the given open subgroup). 2. Show that a subgroup of a Lie group is open if and only if it contains a neighborhood of the identity element.

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3. Conclude that a Lie group is connected if and only if it is generated by arbitrarily small neighborhoods of the identity element. From Proposition 5.1.6 and the fact that the inner automorphisms are group homomorphisms, we get the following corollary. 5.1.8 Corollary. (i) exp(Adg ξ) = g(exp ξ)g −1 , for every ξ ∈ g and g ∈ G; and (ii) Adg [ξ, η] = [Adg ξ, Adg η]. More Automatic Smoothness Results. There are some interesting results related in spirit to Proposition 5.1.4 and the preceding discussions. A striking example of this is the following result: 5.1.9 Theorem.

Any continuous homomorphism of finite-dimensional Lie groups is smooth.

There is a remarkable consequence of this theorem. If G is a topological group (that is, the multiplication and inversion maps are continuous), one could, in principle, have more than one differentiable manifold structure making G into two nonisomorphic Lie groups (i.e., the manifold structures are not diffeomorphic) but both inducing the same topological structure. This phenomenon of “exotic structures” occurs for general manifolds. However, in view of the theorem above, this cannot happen in the case of Lie groups. Indeed, since the identity map is a homeomorphism, it must be a diffeomorphism. Thus, a topological group that is locally Euclidean (i.e., there is an open neighborhood of the identity homeomorphic to an open ball in Rn ) admits at most one smooth manifold structure relative to which it is a Lie group. The existence part of this statement is Hilbert’s famous fifth problem: Show that a locally Euclidean topological group admits a smooth (actually analytic) structure making it into a Lie group. The solution of this problem was achieved by Gleason and, independently, by Montgomery and Zippin in 1952; see Kaplansky [1971] for an excellent account of this proof. Abelian Lie Groups. Since any two elements of an Abelian Lie group G commute, it follows that all adjoint operators Adg , g ∈ G, equal the identity. Therefore, by equation (5.1.5), the Lie algebra g is Abelian; that is, [ξ, η] = 0 for all ξ, η ∈ g.

Examples (a) Any finite-dimensional vector space, thought of as an Abelian group under addition, is an Abelian Lie group. The same is true in infinite dimensions for any Banach space. The exponential map is the identity.  (b) The unit circle in the complex plane S 1 = { z ∈ C | |z| = 1 } is an Abelian Lie group under multiplication. The tangent space Te S 1 is the imaginary axis, and we identify R with Te S 1 by t 7→ 2πit. With this identification, the exponential map exp : R → S 1 is given by exp(t) = e2πit . Note that exp−1 (1) = Z.  (c) The n-dimensional torus Tn = S 1 × · · · × S 1 (n times) is an Abelian Lie group. The exponential map exp : Rn → Tn is given by exp(t1 , . . . , tn ) = (e2πit1 , . . . , e2πitn ). Since S 1 = R/Z , it follows that

Tn = Rn /Zn ,

the projection Rn → Tn being given by exp above.



If G is a connected Lie group whose Lie algebra g is Abelian, the Lie group homomorphism g ∈ G 7→ Adg ∈ GL(g) has induced Lie algebra homomorphism ξ ∈ g 7→ adξ ∈ gl(g) the constant map equal to zero. Therefore, by Corollary 5.1.7, Adg = identity on G, for any g ∈ G. Apply Corollary 5.1.7 again, this time to

5.1 Basic Definitions and Properties

305

the conjugation by g on G (whose induced Lie algebra homomorphism is Adg ), to conclude that it equals the identity map on G. Thus, g commutes with all elements of G; since g was arbitrary, we conclude that G is Abelian. We summarize these observations in the following proposition. 5.1.10 Proposition. If G is an Abelian Lie group, its Lie algebra g is also Abelian. Conversely, if G is connected and g is Abelian, then G is Abelian. The main structure theorem for Abelian Lie groups is the following, whose proof can be found in Varadarajan [1974], or Knapp [2002]. 5.1.11 Theorem. Every connected Abelian n-dimensional Lie group G is isomorphic to a cylinder, that is, to Tk × Rn−k for some k = 1, . . . , n. Lie Subgroups.

It is natural to synthesize the subgroup and submanifold concepts.

5.1.12 Definition. A Lie subgroup H of a Lie group G is a subgroup of G that is also an injectively immersed submanifold of G. If H is a submanifold of G, then H is called a regular Lie subgroup. For example, the one-parameter subgroups of the torus T2 that wind densely on the torus are Lie subgroups that are not regular. The Lie algebras g and h of G and a Lie subgroup H, respectively, are related in the following way: 5.1.13 Proposition.

Let H be a Lie subgroup of G. Then h is a Lie subalgebra of g. Moreover, h = { ξ ∈ g | exp tξ ∈ H for all t ∈ R }.

Proof. The first statement is a consequence of Proposition 5.1.6, which also shows that exp tξ ∈ H, for all ξ ∈ h and t ∈ R. Conversely, if exp tξ ∈ H, for all t ∈ R, we have, d ∈ h, exp tξ dt t=0 since H is a Lie subgroup; but this equals ξ by definition of the exponential map.



The following is a powerful theoretical theorem that can be used to find Lie subgroups. Of course in many examples, such as the orthogonal group, one can readily prove this sort of thing directly using submersion theory. 5.1.14 Theorem. If H is a closed subgroup of a Lie group G, then H is a regular Lie subgroup. Conversely, if H is a regular Lie subgroup of G, then H is closed. The proof of this theorem may be found in Abraham and Marsden [1978], Adams [1969], Varadarajan [1974], or Knapp [2002]. The next result is sometimes called “Lie’s third fundamental theorem.” 5.1.15 Theorem. Let G be a Lie group with Lie algebra g, and let h be a Lie subalgebra of g. Then there exists a unique connected Lie subgroup H of G whose Lie algebra is h. The proof may be found in Knapp [2002] or Varadarajan [1974]. We remind the reader that the Lie algebras appropriate to fluid dynamics and plasma physics are infinitedimensional. Nevertheless, there is still, with the appropriate technical conditions, a correspondence between Lie groups and Lie algebras analogous to the preceding theorems. The reader should be warned, however, that these theorems do not naively generalize to the infinite-dimensional situation, and to prove them for special cases, specialized analytical theorems may be required.

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Quotients. If H is a closed subgroup of G, we denote by G/H, the set of left cosets, that is, the collection { gH | g ∈ G }. Let π : G → G/H be the projection g 7→ gH. 5.1.16 Theorem. There is a unique manifold structure on G/H such that the projection π : G → G/H is a smooth surjective submersion. (Recall from Chapter 4 that a smooth map is called a submersion when its derivative is surjective.) Again the proof may be found in Abraham and Marsden [1978], Knapp [2002], or Varadarajan [1974]. One calls the manifold G/H a homogeneous space.

Exercises  5.1-1.

Verify that Adg [ξ, η] = [Adg ξ, Adg η] directly for GL(n).

 5.1-2. Let G be a Lie group with group operations µ : G × G → G and I : G → G. Show that the tangent bundle T G is also a Lie group, called the tangent group of G with group operations T µ : T G × T G → T G, T I : T G → T G.  5.1-3 (Defining a Lie group by a chart at the identity). Let G be a group and suppose that ϕ : U → V is a one-to-one map from a subset U of G containing the identity element to an open subset V in a Banach space (or Banach manifold). The following conditions are necessary and sufficient for ϕ to be a chart in a Hausdorff–Banach–Lie group structure on G: (a) The set W = { (x, y) ∈ V × V | ϕ−1 (y) ∈ U } is open in V × V , and the map (x, y) ∈ W 7→ ϕ(ϕ−1 (x)ϕ−1 (y)) ∈ V is smooth. (b) For every g ∈ G, the set Vg = ϕ(gU g −1 ∩ U ) is open in V and the map x ∈ Vg 7→ ϕ(gϕ−1 (x)g −1 ) ∈ V is smooth.  5.1-4 (The Heisenberg group). Let (Z, Ω) be a symplectic vector space. That is, Z is a real vector space and Ω : Z × Z → R is a skew symmetric, nondegenerate, bilinear mapping. Let H := Z × S 1 and define the following operation:
(u, exp iφ)(v, exp iψ) = u + v, exp i[φ + ψ + ~−1 Ω(u, v)] . (a) Verify that this operation gives H the structure of a Lie group. (b) Show that the Lie algebra of H is given by h = Z × R with the bracket operation2 [(u, φ), (v, ψ)] = (0, 2~−1 Ω(u, v)). (c) Show that [h, [h, h]] = 0, that is, h is nilpotent, and that R lies in the center of the algebra (i.e., [h, R] = 0); one says that h is a central extension of Z.  5.1-5. Let G be a Lie group and let φ : G → G be the diffeomorphism defined by φ(g) = g −1 . Prove that if X is a left invariant vector field on G then φ∗ (X) is a right invariant vector field whose value at e is −X(e). Further show that X 7→ φ∗ (X) gives a Lie algebra isomorphism of the Lie algebra of left invariant vector fields with the Lie algebra of right invariant vector fields on G. (A Lie algebra isomorphism is a linear mapping A : V → V that preserves the Lie bracket: A([ξ, η]) = [A(ξ), A(η)].) formula for the bracket, when applied to the space Z = R2n of the usual p’s and q’s , shows that this algebra is the same as that encountered in elementary quantum mechanics via the Heisenberg commutation relations. Hence the name “Heisenberg group.” 2 This

5.2 Some Classical Lie Groups

5.2

307

Some Classical Lie Groups

The Real General Linear Group GL(n, R). In the previous section we showed that GL(n, R) is a Lie group, that it is an open subset of the vector space of all linear maps of Rn into itself, and that its Lie algebra is gl(n, R) with the commutator bracket. Since it is open in L(Rn , Rn ) = gl(n, R), the group GL(n, R) is not compact. The determinant function det : GL(n, R) → R is smooth and maps GL(n, R) onto the two components of R\{0}. Thus, GL(n, R) is not connected. Define GL+ (n, R) = { A ∈ GL(n, R) | det(A) > 0 } and note that it is an open (and hence closed) subgroup of GL(n, R). If GL− (n, R) = { A ∈ GL(n, R) | det(A) < 0 }, the map A ∈ GL+ (n, R) 7→ I0 A ∈ GL− (n, R), where I0 is the diagonal matrix all of whose entries are 1 except the (1, 1)-entry, which is −1, is a diffeomorphism. We will show below that GL+ (n, R) is connected, which will prove that GL+ (n, R) is the connected component of the identity in GL(n, R) and that GL(n, R) has exactly two connected components. To prove this we need a theorem from linear algebra called the polar decomposition theorem. To formulate it, recall that a matrix R ∈ GL(n, R) is orthogonal if RRT = RT R = I. A matrix S ∈ gl(n, R) is called symmetric if S T = S. A symmetric matrix S is called positive definite, denoted by S > 0, if hSv, vi > 0 for all v ∈ Rn , v 6= 0. Note that S > 0 implies that S is invertible. 5.2.1 Proposition (Real Polar Decomposition Theorem). For any A ∈ GL(n, R) there exists a unique orthogonal matrix R and positive definite matrices S1 , S2 , such that A = RS1 = S2 R.

(5.2.1)

Proof. Recall first that any positive definite symmetric matrix has a unique square root: If λ1 , . . . , λn > 0 are the eigenvalues of AT A, diagonalize AT A by writing AT A = B diag(λ1 , . . . , λn )B −1 , and then define √ Let S1 =

√

p p AT A = B diag( λ1 , . . . , λn )B −1 .

AT A, which is positive definite and symmetric. Define R = AS1−1 and note that RT R = S1−1 AT AS1−1 = I,

since S12 = AT A by definition. Since both A and S1 are invertible, it follows that R is invertible and hence RT = R−1 , so R is an orthogonal matrix. ˜ S˜1 , then Let us prove uniqueness of the decomposition. If A = RS1 = R ˜ S˜1 = S˜12 . AT A = S1 RT R ˜ = R. However, the square√root of a positive definite matrix is unique, so S1 = S˜1 , whence also R Now define S2 = AAT , and as before, we conclude that A = S2 R0 for some orthogonal matrix R0 . We prove now that R0 = R. Indeed, A = S2 R0 = (R0 (R0 )T )S2 R0 = R0 ((R0 )T S2 R0 ) and (R0 )T S2 R0 > 0. By uniqueness of the prior polar decomposition, we conclude that R0 = R and (R0 )T S2 R0 = S1 . 

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Now we will use the real polar decomposition theorem to prove that GL+ (n, R) is connected. Let A ∈ GL+ (n, R) and decompose it as A = SR, with S positive definite and R an orthogonal matrix whose determinant is 1. We will prove later that the collection of all orthogonal matrices having determinant equal to 1 is a connected Lie group. Thus there is a continuous path R(t) of orthogonal matrices having determinant 1 such that R(0) = I and R(1) = R. Next, define the continuous path of symmetric matrices S(t) = I + t(S − I) and note that S(0) = I and S(1) = S. Moreover, hS(t)v, vi = h[I + t(S − I)]v, vi = kvk2 + thSv, vi − tkvk2 = (1 − t)kvk2 + thSv, vi > 0, for all t ∈ [0, 1], since hSv, vi > 0 by hypothesis. Thus S(t) is a continuous path of positive definite matrices connecting I to S. We conclude that A(t) := S(t)R(t) is a continuous path of matrices whose determinant is strictly positive connecting A(0) = S(0)R(0) = I to A(1) = S(1)R(1) = SR = A. Thus, we have proved the following: 5.2.2 Proposition. The group GL(n, R) is a noncompact disconnected n2 -dimensional Lie group whose Lie algebra gl(n, R) consists of all n × n matrices with the bracket [A, B] = AB − BA. The connected component of the identity is GL+ (n, R), and GL(n, R) has two components. The Real Special Linear Group SL(n, R). Let det : L(Rn , Rn ) → R be the determinant map and recall that GL(n, R) = { A ∈ L(Rn , Rn ) | det A 6= 0 }, so GL(n, R) is open in L(Rn , Rn ). Notice that R\{0} is a group under multiplication and that det : GL(n, R) → R\{0} is a Lie group homomorphism because det(AB) = (det A)(det B). 5.2.3 Lemma. The map det : GL(n, R) → R\{0} is C ∞ , and its derivative is given by D detA ·B = (det A) trace(A−1 B). Proof.

The smoothness of det is clear from its formula in terms of matrix elements. Using the identity det(A + λB) = (det A) det(I + λA−1 B),

it suffices to prove d det(I + λC) = trace C. dλ λ=0 This follows from the identity for the characteristic polynomial det(I + λC) = 1 + λ trace C + · · · + λn det C.



Define the real special linear group SL(n, R) by SL(n, R) = { A ∈ GL(n, R) | det A = 1 } = det−1 (1).

(5.2.2)

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309

From Theorem 5.1.14 it follows that SL(n, R) is a closed Lie subgroup of GL(n, R). However, this method invokes a rather subtle result to prove something that is in reality straightforward. To see this, note that it follows from Lemma 5.2.3 that det : GL(n, R) → R is a submersion, so SL(n, R) = det−1 (1) is a smooth closed submanifold and hence a closed Lie subgroup. The tangent space to SL(n, R) at A ∈ SL(n, R) therefore consists of all matrices B such that trace(A−1 B) = 0. In particular, the tangent space at the identity consists of the matrices with trace zero. We have seen that the Lie algebra of GL(n, R) is L(Rn , Rn ) = gl(n, R) with the Lie bracket given by [A, B] = AB − BA. It follows that the Lie algebra sl(n, R) of SL(n, R) consists of the set of n × n matrices having trace zero, with the bracket [A, B] = AB − BA. Since trace(B) = 0 imposes one condition on B, it follows that dim[sl(n, R)] = n2 − 1. In dealing with classical Lie groups it is useful to introduce the following inner product on gl(n, R): hA, Bi = trace(AB T ). Note that kAk2 =

n X

(5.2.3)

a2ij ,

(5.2.4)

i,j=1 2

which shows that this norm on gl(n, R) coincides with the Euclidean norm on Rn . We shall use this norm to show that SL(n, R) is not compact. Indeed, all matrices of the form   1 0 ... t 0 1 . . . 0    .. .. . . ..  . . . . 0

0 ...

1

√

are elements of SL(n, R) whose norm equals n + t2 for any t ∈ R. Thus, SL(n, R) is not a bounded subset of gl(n, R) and hence is not compact. Finally, let us prove that SL(n, R) is connected. As before, we shall use the real polar decomposition theorem and the fact, to be proved later, that the set of all orthogonal matrices having determinant equal to 1 is a connected Lie group. If A ∈ SL(n, R), decompose it as A = SR, where R is an orthogonal matrix having determinant 1 and S is a positive definite matrix having determinant 1. Since S is symmetric, it can be diagonalized, that is, S = B diag(λ1 , . . . , λn )B −1 for some orthogonal matrix B and λ1 , . . . , λn > 0. Define the continuous path ! n−1 Y S(t) = B diag (1 − t) + tλ1 , . . . , (1 − t) + tλn−1 , 1/ ((1 − t) + tλi ) B −1 i=1

for t ∈ [0, 1] and note that by construction, det S(t) = 1; S(t) is symmetric; S(t) is positive definite, since each entry (1 − t) + tλi > 0 for t ∈ [0, 1]; and S(0) = I, S(1) = S. Now let R(t) be a continuous path of orthogonal matrices of determinant 1 such that R(0) = I and R(1) = R. Therefore, A(t) = S(t)R(t) is a continuous path in SL(n, R) satisfying A(0) = I and A(1) = SR = A, thereby showing that SL(n, R) is connected. 5.2.4 Proposition. The Lie group SL(n, R) is a noncompact connected (n2 − 1)-dimensional Lie group whose Lie algebra sl(n, R) consists of the n × n matrices with trace zero (or linear maps of Rn to Rn with trace zero) with the bracket [A, B] = AB − BA.

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The Orthogonal Group O(n).

On Rn we use the standard inner product hx, yi =

n X

xi y i ,

i=1 1

n

where x = (x , . . . , x ) ∈ R orthogonal if

n

1

n

and y = (y , . . . , y ) ∈ Rn . Recall that a linear map A ∈ L(Rn , Rn ) is hAx, Ayi = hx, yi

(5.2.5)

1/2

for all x, y ∈ R. In terms of the norm kxk = hx, xi , one sees from the polarization identity that A is n T orthogonal

 iff kAxk = kxk, for all x ∈ R , orT in terms of the transpose A , which is defined by hAx, yi = T x, A y , we see that A is orthogonal iff AA = I. Let O(n) denote the orthogonal elements of L(Rn , Rn ). For A ∈ O(n), we see that 1 = det(AAT ) = (det A)(det AT ) = (det A)2 ; hence det A = ±1, and so A ∈ GL(n, R). Furthermore, if A, B ∈ O(n), then hABx, AByi = hBx, Byi = hx, yi , and so AB ∈ O(n). Letting x0 = A−1 x and y0 = A−1 y, we see that hx, yi = hAx0 , Ay0 i = hx0 , y0 i , that is,

 hx, yi = A−1 x, A−1 y ; hence A−1 ∈ O(n). Let S(n) denote the vector space of symmetric linear maps of Rn to itself, and let ψ : GL(n, R) → S(n) be defined by ψ(A) = AAT . We claim that I is a regular value of ψ. Indeed, if A ∈ ψ −1 (I) = O(n), the derivative of ψ is Dψ(A) · B = AB T + BAT , which is onto (to hit C, take B = CA/2). Thus, ψ −1 (I) = O(n) is a closed Lie subgroup of GL(n, R), called the orthogonal group. The group O(n) is also bounded in L(Rn , Rn ): The norm of A ∈ O(n) is  1/2 √ 1/2 kAk = trace(AT A) = (trace I) = n. Therefore, O(n) is compact. We shall see in §9.3 that O(n) is not connected, but has two connected components, one where det = +1 and the other where det = −1. The Lie algebra o(n) of O(n) is ker Dψ(I), namely, the skew-symmetric linear maps with the usual commutator bracket [A, B] = AB − BA. The space of skew-symmetric n × n matrices has dimension equal to the number of entries above the diagonal, namely, n(n − 1)/2. Thus, dim[O(n)] = 21 n(n − 1). The special orthogonal group is defined as SO(n) = O(n) ∩ SL(n, R), that is, SO(n) = { A ∈ O(n) | det A = +1 }.

(5.2.6) −1

Since SO(n) is the kernel of det : O(n) → {−1, 1}, that is, SO(n) = det (1), it is an open and closed Lie subgroup of O(n), hence is compact. We shall prove in §9.3 that SO(n) is the connected component of O(n) containing the identity I, and so has the same Lie algebra as O(n). We summarize: 5.2.5 Proposition. The Lie group O(n) is a compact Lie group of dimension n(n − 1)/2. Its Lie algebra o(n) is the space of skew-symmetric n×n matrices with bracket [A, B] = AB −BA. The connected component of the identity in O(n) is the compact Lie group SO(n), which has the same Lie algebra so(n) = o(n). The Lie group O(n) has two connected components.

5.2 Some Classical Lie Groups

Rotations in the Plane SO(2).

311

We parametrize S 1 = { x ∈ R2 | kxk = 1 }

by the polar angle θ, 0 ≤ θ < 2π. For each θ ∈ [0, 2π], let   cos θ − sin θ Aθ = , sin θ cos θ using the standard basis of R2 . Then Aθ ∈ SO(2) represents a counter-clockwise rotation through the angle θ. Conversely, if   a1 a2 A= a3 a4 is in SO(2), the relations a21 + a22 = 1, a23 + a24 = 1, a1 a3 + a2 a4 = 0, det A = a1 a4 − a2 a3 = 1 show that A = Aθ for some θ. Thus, SO(2) can be identified with S 1 , that is, with rotations in the plane. Rotations in Space SO(3). The Lie algebra so(3) of SO(3) may be define the vector space isomorphism ˆ : R3 → so(3), called the hat map,  0 −v3 v2 ˆ =  v3 0 −v1 v = (v1 , v2 , v3 ) 7→ v −v2 v1 0

identified with R3 as follows. We by  .

(5.2.7)

Note that the identity ˆw = v × w v characterizes this isomorphism. We get ˆ−v ˆu ˆ) w = u ˆ (v × w) − v ˆ (u × w) (ˆ uv = u × (v × w) − v × (u × w) = (u × v) × w = (u × v)ˆ · w. Thus, if we put the cross product on R3 , ˆ becomes a Lie algebra isomorphism, and so we can identify so(3) with R3 carrying the cross product as Lie bracket. We also note that the standard dot product may be written
ˆT w ˆ = − 21 trace (ˆ ˆ . vw) v · w = 21 trace v 5.2.6 Theorem (Euler’s Theorem). about an axis w.

Every element A ∈ SO(3), A 6= I, is a rotation through an angle θ

To prove this, we use the following lemma: 5.2.7 Lemma.

Every A ∈ SO(3) has an eigenvalue equal to 1.

Proof. The eigenvalues of A are given by roots of the third-degree polynomial det(A − λI) = 0. Roots occur in conjugate pairs, so at least one is real. If λ is a real root and x is a nonzero real eigenvector, then Ax = λx, so kAxk2 = kxk2 and kAxk2 = |λ|2 kxk2 imply λ = ±1. If all three roots are real, they are (1, 1, 1) or (1, −1, −1), since det A = 1. If there is one real and two complex conjugate roots, they are (1, ω, ω ¯ ), since det A = 1. In any case, one real root must be +1. 

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Proof of Theorem 5.2.6. By Lemma 5.2.7, the matrix A has an eigenvector w with eigenvalue 1, say Aw = w. The line spanned by w is also invariant under A. Let P be the plane perpendicular to w; that is, P = { y | hw, yi = 0 } . Since A is orthogonal, A(P ) = P . Let e1 , e2 be an orthogonal basis in P . Then relative to (w, e1 , e2 ), A has the matrix   1 0 0 A =  0 a1 a2  . 0 a3 a4 Since



a1 a3

a2 a4



lies in SO(2), A is a rotation about the axis w by some angle. 5.2.8 Corollary.



Any A ∈ SO(3) can be written in some orthonormal basis as the matrix   1 0 0 A =  0 cos θ − sin θ  . 0 sin θ cos θ

The infinitesimal version of Euler’s theorem is the following: ˆ is a 5.2.9 Proposition. Identifying the Lie algebra so(3) of SO(3) with the Lie algebra R3 , exp(tw) rotation about w by the angle tkwk, where w ∈ R3 . Proof. To simplify the computation, we pick an orthonormal basis {e1 , e2 , e3 } of R3 , with e1 = w/kwk. ˆ has the matrix Relative to this basis, w   0 0 0 ˆ = kwk  0 0 −1  . w 0 1 0 Let 1 c(t) =  0 0 

0 cos tkwk sin tkwk

 0 − sin tkwk  . cos tkwk

Then 0 c0 (t) =  0 0 

 0 0 −kwk sin tkwk −kwk cos tkwk  kwk cos tkwk −kwk sin tkwk

ˆ = TI Lc(t) (w) ˆ = Xwˆ (c(t)), = c(t)w ˆ Therefore, c(t) is an integral curve of Xwˆ ; where Xwˆ is the left-invariant vector field corresponding to w. ˆ is also an integral curve of Xwˆ . Since both agree at t = 0, exp(tw) ˆ = c(t), for all t ∈ R. But but exp(tw) the matrix definition of c(t) expresses it as a rotation by an angle tkwk about the axis w.  Despite Euler’s theorem, it might be good to recall now that SO(3) cannot be written as S 2 × S 1 . Amplifying on Proposition 5.2.9, we give the following explicit formula for exp ξ, where ξ ∈ so(3), which is called Rodrigues’ formula:   2  kvk sin 2 sin kvk  v ˆ2. ˆ + 12  v (5.2.8) exp[ˆ v] = I + kvk kvk 2

5.2 Some Classical Lie Groups

313

This formula was given by Rodrigues in 1840; see also Exercise 1 in Helgason [2001] and see Altmann [1986] for some interesting history of this formula. Proof of Rodrigues’ Formula.

By (5.2.7),

ˆ 2 w = v × (v × w) = hv, wi v − kvk2 w. v

(5.2.9)

Consequently, we have the recurrence relations ˆ 3 = −kvk2 v ˆ, v

ˆ 4 = −kvk2 v ˆ2, v

ˆ 5 = kvk4 v ˆ, v

ˆ 6 = kvk4 v ˆ2, . . . . v

Splitting the exponential series in odd and even powers,   kvk2 kvk4 kvk2n ˆ exp[ˆ v] = I + I − + − · · · + (−1)n+1 + ··· v 3! 5! (2n + 1)!   1 kvk2 kvk4 kvkn−2 ˆ2 + − + + · · · + (−1)n−1 + ··· v 2! 4! 6! (2n)! sin kvk 1 − cos kvk 2 ˆ+ ˆ , =I+ v v kvk kvk2 and so the result follows from the identity 2 sin2 (kvk/2) = 1 − cos kvk.

(5.2.10) 

The following alternative expression, equivalent to (5.2.8), is often useful. Set n = v/kvk, so that knk = 1. From (5.2.9) and (5.2.10) we obtain exp[ˆ v] = I + (sin kvk)ˆ n + (1 − cos kvk)[n ⊗ n − I].

(5.2.11)

Here, n ⊗ n is the matrix whose entries are ni nj , or as a bilinear form, (n ⊗ n)(α, β) = n(α)n(β). Therefore, we obtain a rotation about the unit vector n = v/kvk of magnitude kvk. The results (5.2.8) and (5.2.11) are useful in computational solid mechanics, along with their quaternionic counterparts. We shall return to this point below in connection with SU(2); see Whittaker [1988] and Simo and Fox [1989] for more information. We next give a topological property of SO(3). 5.2.10 Proposition.

The rotation group SO(3) is diffeomorphic to the real projective space RP3 .

Proof. To see this,pmap the unit ball D in R3 to SO(3) by sending (x, y, z) to the rotation about (x, y, z) through the angle π x2 + y 2 + z 2 (and (0, 0, 0) to the identity). This mapping is clearly smooth and surjective. Its restriction to the interior of D is injective. On the boundary of D, this mapping is 2 to 1, so it induces a smooth bijective map from D, with antipodal points on the boundary identified, to SO(3). It is a straightforward exercise to show that the inverse of this map is also smooth. Thus, SO(3) is diffeomorphic with D, with antipodal points on the boundary identified. However, the mapping p (x, y, z) 7→ (x, y, z, 1 − x2 − y 2 − z 2 ) is a diffeomorphism between D, with antipodal points on the boundary identified, and the upper unit hemisphere of S 3 with antipodal points on the equator identified. The latter space is clearly diffeomorphic to the unit sphere S 3 with antipodal points identified, which coincides with the space of lines in R4 through the origin, that is, with RP3 . 

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The Real Symplectic Group Sp(2n, R).

Let 

0 J= −I

 I . 0

A linear operator A ∈ L(R2n , R2n ) is called symplectic if AT JA = J. Let Sp(2n, R) be the set of 2n × 2n symplectic matrices. Taking determinants of the condition AT JA = J gives 1 = det J = (det AT ) · (det AJ) · (det A) = (det A)2 . Hence, det A = ±1, and so A ∈ GL(2n, R). Furthermore, if A, B ∈ Sp(2n, R), then (AB)T J(AB) = B T AT JAB = J. Hence, AB ∈ Sp(2n, R), and if AT JA = J, then JA = (AT )−1 J = (A−1 )T J, so J = A−1


T

JA−1 ,

or A−1 ∈ Sp(2n, R).

Thus, Sp(2n, R) is a group. If A=

  a b ∈ GL(2n, R), c d

then (see Exercise 2.3-2) ( aT c and bT d are symmetric and A ∈ Sp(2n, R) iff aT d − cT b = 1.

(5.2.12)

Define ψ : GL(2n, R) → so(2n) by ψ(A) = AT JA. Let us show that J is a regular value of ψ. Indeed, if A ∈ ψ −1 (J) = Sp(2n, R), the derivative of ψ is Dψ(A) · B = B T JA + AT JB. Now, if C ∈ so(2n), let B = − 21 AJC. We verify, using the identity AT J = JA−1 , that Dψ(A) · B = C. Indeed, B T JA + AT JB = B T (A−1 )T J + JA−1 B = (A−1 B)T J + J(A−1 B) = (− 21 JC)T J + J(− 12 JC) = − 21 C T JT J − 12 J2 C = − 21 CJ2 − 12 J2 C = C, since JT = −J and J2 = −I. Thus Sp(2n, R) = ψ −1 (J) is a closed smooth submanifold of GL(2n, R) whose Lie algebra is 
ker Dψ(J) = B ∈ L R2n , R2n | B T J + JB = 0 .

5.2 Some Classical Lie Groups

315

The Lie group Sp(2n, R) is called the symplectic group, and its Lie algebra
 sp(2n, R) = A ∈ L R2n , R2n | AT J + JA = 0 the symplectic algebra. Moreover, if A=

  a b ∈ sl(2n, R), c d

then A ∈ sp(2n, R) iff d = −aT , c = cT , and b = bT .

(5.2.13)

2

The dimension of sp(2n, R) can be readily calculated to be 2n + n. Using (5.2.12), it follows that all matrices of the form   I 0 tI I √ are symplectic. However, the norm of such a matrix is equal to 2n + t2 n, which is unbounded if t ∈ R. Therefore, Sp(2n, R) is not a bounded subset of gl(2n, R) and hence is not compact. We next summarize what we have found. 5.2.11 Proposition.

The symplectic group Sp(2n, R) := { A ∈ GL(2n, R) | AT JA = J }

is a noncompact, connected Lie group of dimension 2n2 + n. Its Lie algebra sp(2n, R) consists of the 2n × 2n matrices A satisfying AT J + JA = 0, where   0 I J= −I 0 with I the n × n identity matrix. We shall indicate in the next section how one proves that Sp(2n, R) is connected. We now prove that symplectic linear maps must have determinant 1 (this is notably different from the case of orthogonal transformations which may have determinant either 1 or −1. 5.2.12 Lemma.

If A ∈ Sp(n, R), then det A = 1.

Proof. Since AT JA = J and det J = 1, it follows that (det A)2 = 1. Unfortunately, this still leaves open the possibility that det A = −1. To eliminate it, we proceed in the following way. Define the symplectic form Ω on R2n by Ω(u, v) = uT Jv, that is, relative to the chosen basis of R2n , the matrix of Ω is J. As we know from linear algebra, the volume of the parallelopiped spanned by vectors v1 , . . . , v2n in R2n is given by µ(v1 , . . . , v2n ) = det (Ω(vi , vj )) . But the determinant of a linear map measures how volumes are tranformed; that is, (det A)µ = A∗ µ, and so we get (det A)µ (v1 , . . . , v2n ) = (A∗ µ) (v1 , . . . , v2n ) = µ (Av1 , . . . , Av2n ) = det (Ω (Avi , Avj )) = det (Ω (vi , vj )) = µ (v1 , . . . , v2n ) , since A ∈ Sp(2n, R), which is equivalent to Ω(Au, Av) = Ω(u, v) for all u, v ∈ R2n . Taking v1 , . . . , v2n to be the standard basis of R2n , we conclude that det A = 1. 

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5.2.13 Proposition (Symplectic Eigenvalue Theorem). If λ0 ∈ C is an eigenvalue of A ∈ Sp(2n, R) of multiplicity k, then 1/λ0 , λ0 , and 1/λ0 are eigenvalues of A of the same multiplicity k. Moreover, if ±1 occur as eigenvalues, their multiplicities are even. Proof. Since A is a real matrix, if λ0 is an eigenvalue of A of multiplicity k, so is λ0 by elementary algebra. Let us show that 1/λ0 is also an eigenvalue of A. If p(λ) = det(A − λI) is the characteristic polynomial of A, since
T JAJ−1 = A−1 , det J = 1, J−1 = −J = JT , and det A = 1 (by the Proposition just proved), we get   p(λ) = det(A − λI) = det J(A − λI)J−1 
T  = det(JAJ−1 − λI) = det A−1 − λI
= det(A−1 − λI) = det A−1 (I − λA)    1 = det(I − λA) = det λ I −A λ   1 = λ2n det I −A λ   1 = λ2n (−1)2n det A − I λ   1 = λ2n p . λ

(5.2.14)

Since 0 is not an eigenvalue of A, it follows that p(λ) = 0 iff p (1/λ) = 0, and hence, λ0 is an eigenvalue of A iff 1/λ0 is an eigenvalue of A. Now assume that λ0 has multiplicity k, that is, p(λ) = (λ − λ0 )k q(λ) for some polynomial q(λ) of degree 2n − k satisfying q(λ0 ) 6= 0. Since p(λ) = λ2n p(1/λ), we conclude that p(λ) = p

1 λ




λ2n = (λ − λ0 )k q(λ) = (λλ0 )k



1 1 − λ0 λ

k q(λ).

However, λk0 q(λ) 2n−k λ is a polynomial in 1/λ, since the degree of q(λ) is 2n − k, k ≤ 2n. Thus 1/λ0 is a root of p(λ) having multiplicity l ≥ k. Reversing the roles of λ0 and 1/λ0 , we similarly conclude that k ≥ l, and hence it follows that k = l. Finally, note that λ0 = 1/λ0 iff λ0 = ±1. Thus, since all eigenvalues of A occur in pairs whose product is 1 and the size of A is 2n × 2n, it follows that the total number of times +1 and −1 occur as eigenvalues is even. However, since det A = 1, we conclude that −1 occurs an even number of times as an eigenvalue of A (if it occurs at all). Therefore, the multiplicity of 1 as an eigenvalue of A, if it occurs, is also even.  Figure 5.2.1 illustrates the possible configurations of the eigenvalues of A ∈ Sp(4, R). Next, we study the eigenvalues of matrices in sp(2n, R). The following theorem is useful in the stability analysis of relative equilibria. If A ∈ sp(2n, R), then AT J + JA = 0, so that if p(λ) = det(A − λI) is the

5.2 Some Classical Lie Groups

y

317

y

x

x

saddle center y

complex saddle y

x

real saddle y

x

generic center y

y

x

degenerate saddle

x

x

identity

degenerate center

Figure 5.2.1. Symplectic eigenvalue theorem on R4 .

characteristic polynomial of A, we have p(λ) = det(A − λI) = det(J(A − λI)J) = det(JAJ + λI) = det(−AT J2 + λI) = det(AT + λI) = det(A + λI) = p(−λ). In particular, notice that trace(A) = 0. Proceeding as before and using this identity, we conclude the following: 5.2.14 Proposition (Infinitesimally Symplectic Eigenvalues). If λ0 ∈ C is an eigenvalue of A ∈ sp(2n, R) of multiplicity k, then −λ0 , λ0 , and −λ0 are eigenvalues of A of the same multiplicity k. Moreover, if 0 is an eigenvalue, it has even multiplicity. Figure 9.2.2 shows the possible infinitesimally symplectic eigenvalue configurations for A ∈ sp(4, R). The Symplectic Group and Mechanics. Consider a particle of mass m moving in a potential V (q), where q = (q 1 , q 2 , q 3 ) ∈ R3 . Newton’s second law states that the particle moves along a curve q(t) in R3 in such a way that m¨ q = − grad V (q). Introduce the momentum pi = mq˙i , i = 1, 2, 3, and the energy 3

H(q, p) = Then

1 X 2 p + V (q). 2m i=1 i

∂H ∂V = i = −m¨ qi = −p˙i , ∂q i ∂q

and

∂H 1 = pi = q˙i , ∂pi m

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5. An Introduction to Lie Groups

y

y

x

x

saddle center y

complex saddle y

x

x

real saddle y

generic center y

y

x

x

x

degenerate saddle

identity

degenerate center

Figure 5.2.2. Infinitesimally symplectic eigenvalue theorem on R4 .

and hence Newton’s law F = ma is equivalent to Hamilton’s equations q˙i =

∂H , ∂pi

p˙i = −

∂H , ∂q i

i = 1, 2, 3.

Writing z = (q, p),  ∂H  ∂q    ˙ p) ˙ = z, ˙  ∂H  = (q, ∂p   J · grad H(z) =

0 −I

I 0



so Hamilton’s equations read z˙ = J · grad H(z). For two functions F, K : R2n → R, their Poisson Bracket is defined to be {F, K} (z) = (grad F )T J grad K =

 n  X ∂F ∂K ∂K ∂F − ∂q i ∂pi ∂q i ∂pi i=1

(5.2.15)

From Hamilton’s equations in the form z˙ = J·grad H(z) together with the chain rule, we see that Hamilton’s equations can be equivalently written in Poisson bracket form: F˙ = {F, H} for all smooth functions F . Next we turn to the transformation properties of Hamilton’s equations. Let f : R3 × R 3 → R3 × R 3

5.2 Some Classical Lie Groups

319

and write w = f (z). If z(t) satisfies Hamilton’s equations z˙ = J · grad H(z), then w(t) = f (z(t)) satisfies w˙ = AT z, ˙ where AT = [∂wi /∂z j ] is the Jacobian matrix of f . By the chain rule, w˙ = AT J gradz H(z) = AT JA gradw H(z(w)). Thus, the equations for w(t) have the form of Hamilton’s equations with energy K(w) = H(z(w)) if and only if AT JA = J, that is, iff A is symplectic. A nonlinear transformation f is canonical iff its Jacobian matrix is symplectic. As a special case, consider a linear map A ∈ Sp(2n, R) and let w = Az. Suppose H is quadratic, that is, of the form H(z) = hz, Bzi/2, where B is a symmetric 2n × 2n matrix. Then grad H(z) · δz = =

1 2 hδz, Bzi + hz, Bδzi 1 2 (hδz, Bzi + hBz, δzi)

= hδz, Bzi ,

so grad H(z) = Bz and thus the equations of motion become the linear equations z˙ = JBz. Now w˙ = Az˙ = AJBz = J(AT )−1 Bz = J(AT )−1 BA−1 Az = JB 0 w, where B 0 = (AT )−1 BA−1 is symmetric. For the new Hamiltonian we get



 H 0 (w) = 21 w, (AT )−1 BA−1 w = 12 A−1 w, BA−1 w = H(A−1 w) = H(z). Thus, Sp(2n, R) is the linear invariance group of classical mechanics. The Complex General Linear Group GL(n, C). Many important Lie groups involve complex matrices. As in the real case, GL(n, C) = { n × n invertible complex matrices } is an open set in L(Cn , Cn ) = { n × n complex matrices }. Clearly, GL(n, C) is a group under matrix multiplication. Therefore, GL(n, C) is a Lie group and has the Lie algebra gl(n, C) = { n × n complex matrices } = L(Cn , Cn ). Hence GL(n, C) has complex dimension n2 , that is, real dimension 2n2 . We shall prove below that GL(n, C) is connected (contrast this with the fact that GL(n, R) has two components). As in the real case, we will need a polar decomposition theorem to do this. A matrix U ∈ T GL(n, C) is unitary if U U † = U † U = I, where U † := U . A matrix P ∈ gl(n, C) is called Hermitian if P † = P . A Hermitian matrix P is called positive definite, denoted by P > 0, if hP z, zi > 0 for all z ∈ Cn , z 6= 0, where h , i denotes the inner product on Cn . Note that P > 0 implies that P is invertible. 5.2.15 Proposition (Complex Polar Decomposition). For any matrix A ∈ GL(n, C), there exists a unique unitary matrix U and positive definite Hermitian matrices P1 , P2 such that A = U P1 = P2 U. The proof is identical to that of Proposition 5.2.1 with the obvious changes. The only additional property needed is the fact that the eigenvalues of a Hermitian matrix are real. As in the proof of the real case, one needs to use the connectedness of the space of unitary matrices (proved in the next section) to conclude the following: 5.2.16 Proposition. The group GL(n, C) is a complex noncompact connected Lie group of complex dimension n2 and real dimension 2n2 . Its Lie algebra gl(n, C) consists of all n × n complex matrices with the commutator bracket. On gl(n, C), there is an inner product defined by hA, Bi = trace(AB † ).

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The Complex Special Linear Group.

This group is defined by

SL(n, C) := { A ∈ GL(n, C) | det A = 1 } and is treated as in the real case. In the proof of its connectedness one uses the complex polar decomposition theorem and the fact that any Hermitian matrix can be diagonalized by conjugating it with an appropriate unitary matrix. 5.2.17 Proposition. The group SL(n, C) is a complex noncompact Lie group of complex dimension n2 − 1 and real dimension 2(n2 − 1). Its Lie algebra sl(n, C) consists of all n × n complex matrices of trace zero with the commutator bracket. The Unitary Group U(n).

Recall that Cn has the Hermitian inner product hx, yi =

n X

xi y¯i ,

i=0



where x = x1 , . . . , xn ∈ Cn , y = y 1 , . . . , y n ∈ Cn , and y¯i denotes the complex conjugate. Let U(n) = { A ∈ GL(n, C) | hAx, Ayi = hx, yi } . † † The orthogonality A† = A¯T , that is,

condition hAx, Ayi = hx, yi is equivalent to AA = A A = I, where † 1 hAx, yi = x, A y . From | det A| = 1, we see that det maps U(n) into the unit circle S = { z ∈ C | |z| = 1 }. As is to be expected by now, U(n) is a closed Lie subgroup of GL(n, C) with Lie algebra

u(n) = { A ∈ L(Cn , Cn ) | hAx, yi = − hx, Ayi } = { A ∈ gl(n, C) | A† = −A }; the proof parallels that for O(n). The elements of u(n) are called skew-Hermitian matrices. Since the norm of A ∈ U(n) is
1/2 √ kAk = trace(A† A) = (trace I)1/2 = n, it follows that U(n) is closed and bounded, hence compact, in GL(n, C). From the definition of u(n) it immediately follows that the real dimension of U(n) is n2 . Thus, even though the entries of the elements of U(n) are complex, U(n) is a real Lie group. In the special case n = 1, a complex linear map ϕ : C → C is multiplication by some complex number z, and ϕ is an isometry if and only if |z| = 1. In this way the group U(1) is identified with the unit circle S 1 . The special unitary group SU(n) = { A ∈ U(n) | det A = 1 } = U(n) ∩ SL(n, C) is a closed Lie subgroup of U(n) with Lie algebra su(n) = { A ∈ L(Cn , Cn ) | hAx, yi = − hx, Ayi and trace A = 0 }. Hence, SU(n) is compact and has (real) dimension n2 − 1. We shall prove later that both U(n) and SU(n) are connected. 5.2.18 Proposition. The group U(n) is a compact real Lie subgroup of GL(n, C) of (real ) dimension n2 . Its Lie algebra u(n) consists of the space of skew-Hermitian n × n matrices with the commutator bracket. SU(n) is a closed real Lie subgroup of U(n) of dimension n2 − 1 whose Lie algebra su(n) consists of all trace zero skew-Hermitian n × n matrices. Another related interesting fact (that we just state) is that Sp(2n, R) ∩ O(2n, R) = U(n).

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321

The Group SU(2). This group warrants special attention, since it appears in many physical applications such as the Cayley–Klein parameters for the free rigid body and in the construction of the (nonabelian) gauge group for the Yang–Mills equations in elementary particle physics. From the general formula for the dimension of SU(n) it follows that dim SU(2) = 3. The group SU(2) is diffeomorphic to the three-sphere S 3 = { x ∈ R4 | kxk = 1 }, with the diffeomorphism given by  0  x − ix3 −x2 − ix1 x = (x0 , x1 , x2 , x3 ) ∈ S 3 ⊂ R4 7→ ∈ SU(2). (5.2.16) x2 − ix1 x0 + ix3 Therefore, SU(2) is connected and simply connected. By Euler’s Theorem 5.2.6 every element of SO(3) different from the identity is determined by a vector v, which we can choose to be a unit vector, and an angle of rotation θ about the axis v. The trouble is, the pair (v, θ) and (−v, −θ) represent the same rotation and there is no consistent way to continuously choose one of these pairs, valid for the entire group SO(3). Such a choice is called, in physics, a choice of spin. This suggests the existence of a double cover of SO(3) that, hopefully, should also be a Lie group. We will show below that SU(2) fulfills these requirements3 . This is based on the following construction. Let σ1 , σ2 , σ3 be the Pauli spin matrices, defined by       0 1 0 −i 1 0 σ1 = , σ2 = , and σ3 = , 1 0 i 0 0 −1 and let σ = (σ1 , σ2 , σ3 ). Then one checks that [σ1 , σ2 ] = 2iσ3 (plus cyclic permutations), from which one finds that the map " −ix3 1 1 ˜ = x·σ = x 7→ x 2i 2 −ix1 + x2

−ix1 − x2 ix3

# ,

where x · σ = x1 σ1 + x2 σ2 + x3 σ3 , is a Lie algebra isomorphism between R3 and the 2 × 2 skew-Hermitian ˜ ] = (x × y)˜. Note that traceless matrices (the Lie algebra of SU(2)); that is, [˜ x, y ˜ ) = − 12 x · y. − det(x · σ) = kxk2 , and trace (˜ xy Define the Lie group homomorphism π : SU(2) → GL(3, R) by (π(A)x) · σ = A(x · σ)A† = A(x · σ)A−1 .

(5.2.17)

A straightforward computation, using the expression (5.2.16), shows that ker π = {±I}. Therefore, π(A) = π(B) if and only if A = ±B. Since kπ(A)xk2 = − det((π(A)x) · σ) = − det(A(x · σ)A−1 ) = − det(x · σ) = kxk2 , it follows that π(SU(2)) ⊂ O(3). But π(SU(2)) is connected, being the continuous image of a connected space, and so π(SU(2)) ⊂ SO(3). 3 For any SO(n), it is a theorem that there is a unique simply connected 2:1 covering group, called the spin group and denoted by Spin(n). We shall, in effect, show below that Spin(3) = SU(2).

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˜ ∈ su(2), then Let us show that π : SU(2) → SO(3) is a local diffeomorphism. Indeed, if α ˜ ˜ † + α(x ˜ · σ) (Te π(α)x) · σ = (x · σ)α ˜ x · σ] = 2i[α, ˜ x ˜] = [α, ˜ × x)˜ = (α ˜ × x) · σ = 2i(α ˆ · σ, = (αx) ˜ = α. ˆ Thus, that is, Te π(α) Te π : su(2) −→ so(3) is a Lie algebra isomorphism and hence π is a local diffeomorphism in a neighborhood of the identity. Since π is a Lie group homomorphism, it is a local diffeomorphism around every point. In particular, π(SU(2)) is open and hence closed (its complement is a union of open cosets in SO(3)). Since it is nonempty and SO(3) is connected, we have π(SU(2)) = SO(3). Therefore, π : SU(2) → SO(3) is a 2 to 1 surjective submersion. Summarizing, we have the commutative diagram in Figure 9.2.3.

S3

≈

2:1 ? RP3

- SU(2) 2:1 ? - SO(3)

≈ Figure 5.2.3. The link between SU(2) and SO(3).

5.2.19 Proposition.

The Lie group SU(2) is the simply connected 2 to 1 covering group of SO(3).

Quaternions. The division ring H (or, by abuse of language, the noncommutative field) of quaternions is generated over the reals by three elements i, j, k with the relations i2 = j2 = k2 = −1, ij = −ji = k, jk = −kj = i, ki = −ik = j. Quaternionic multiplication is performed in the usual manner (like polynomial multiplication) taking the above relations into account. If a ∈ H, we write a = (as , av ) = as + a1v i + a2v j + a3v k for the scalar and vectorial part of the quaternion, where as , a1v , a2v , a3v ∈ R. Quaternions having zero scalar part are also called pure quaternions. With this notation, quaternionic multiplication has the expression ab = (as bs − av · bv , as bv + bs av + av × bv ) . In addition, every quaternion a = (as , av ) has a conjugate a := (as , −av ), that is, the real numbers are fixed by the conjugation and i = −i, j = −j, and k = −k. Note that ab = ba. Every quaternion a 6= 0 has an inverse given by a−1 = a/|a|2 , where |a|2 := aa = aa = a2s + kav k2 . In particular, the unit quaternions, which, as a set, equal the unit sphere S 3 in R4 , form a group under quaternionic multiplication.

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5.2.20 Proposition. The unit quaternions S 3 = { a ∈ H | |a| = 1 } form a Lie group isomorphic to SU(2) via the isomorphism (5.2.16).

Proof. We already noted that (5.2.16) is a diffeomorphism of S 3 with SU(2), so all that remains to be shown is that it is a group homomorphism, which is a straightforward computation. 

Since the Lie algebra of S 3 is the tangent space at 1, it follows that it is isomorphic to the pure quaternions R . We begin by determining the adjoint action of S 3 on its Lie algebra. If a ∈ S 3 and bv is a pure quaternion, the derivative of the conjugation is given by 3

a 1 = 2 (−av · bv , as bv + av × bv )(as , −av ) |a|2 |a|
0, 2as (av × bv ) + 2(av · bv )av + (a2s − kav k2 )bv .

Ada bv = abv a−1 = abv =

1 |a|2

Therefore, if a(t) = (1, tav ), we have a(0) = 1, a0 (0) = av , so that the Lie bracket on the pure quaternions R3 is given by d Ada(t) bv dt t=0 1 d (2t(av × bv ) + 2t2 (av · bv )av = dt t=0 1 + t2 kav k2
+ 1 − t2 kav k2 )bv = 2av × bv .

[av , bv ] =

Thus, the Lie algebra of S 3 is R3 relative to the Lie bracket given by twice the cross product of vectors. The derivative of the Lie group isomorphism (5.2.16) is given by −ix3 x∈R → 7 −ix1 + x2 3



 −ix1 − x2 = 2˜ x ∈ su(2), ix3

and is thus a Lie algebra isomorphism from R3 with twice the cross product as bracket to su(2), or equivalently to (R3 , ×). Let us return to the commutative diagram in Figure 9.2.3 and determine explicitly the 2 to 1 surjective map S 3 → SO(3) that associates to a quaternion a ∈ S 3 ⊂ H the rotation matrix A ∈ SO(3). To compute this map, let a ∈ S 3 and associate to it the matrix as − ia3v a2v − ia1v

 U=

 −a2v − ia1v , as + ia3v

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where a = (as , av ) = (as , a1v , a2v , a3v ). By (5.2.17), the rotation matrix is given by A = π(U ), namely, (Ax) · σ = (π(U )x) · σ = U (x · σ)U †    x3 x1 − ix2 as − ia3v −a2v − ia1v = 2 av − ia1v as + ia3v x1 + ix2 −x3   as + ia3v a2v + ia1v × −a2v + ia1v as − ia3v  2
= as + (a1v )2 − (a2v )2 − (a3v )2 x1 + 2(a1v a2v − as a3v )x2  +2(as a2v + a1v a3v )x3 σ1 

+ 2 a1v a2v + as a3v x1 + a2s − (a1v )2 + (a2v )2 − (a3v )2 x2
 +2 a2v a3v − as a1v x3 σ2 

+ 2 a1v a3v − as a2v x1 + 2 as a1v + a2v a3v x2
 + a2s − (a1v )2 − (a2v )2 + (a3v )2 x3 σ3 . Thus, taking into account that a2s + (a1v )2 + (a2v )2 + (a3v )2 = 1, we get the expression of the matrix A as 2a2s + 2(a1v )2 − 1  2(as a3v + a1v a2v ) 2(−as a1v + a2v a3v ) 

2(−as a3v + a1v a2v ) 2a2s + 2(a2v )2 − 1 2(as a1v + a2v a3v )

 2(as a2v + a1v a3v ) 2(−as a1v + a2v a3v ) 2a2s + (a3v )2 − 1 ˆv + 2av ⊗ av , (5.2.18) = (2a2s − 1)I + 2as a

where av ⊗ av is the symmetric matrix whose (i, j) entry equals aiv ajv . The map ˆv + 2av ⊗ av a ∈ S 3 7→ (2a2s − 1)I + 2as a is called the Euler–Rodrigues parametrization. It has the advantage, as opposed to the Euler angles parametrization, which has a coordinate singularity, of being global. This is of crucial importance in computational mechanics (see, for example, Marsden and West [2001]). Finally, let us rewrite Rodrigues’ formula (5.2.8) in terms of unit quaternions. Let  ω  ω  a = (as , av ) = cos , sin n , 2 2 ˆ 2 = n ⊗ n − I, from (5.2.8) we get where ω > 0 is an angle and n is a unit vector. Since n  ω exp(ωn) = I + (sin ω)ˆ n + 2 sin2 (n ⊗ n − I) 2    ω ω ω ω ˆ + 2 sin2 = 1 − 2 sin2 I + 2 cos sin n n⊗n 2 2 2 2
ˆv + 2av ⊗ av . = 2a2s − 1 I + 2as a This expression then produces a rotation associated to each unit quaternion a. In addition, using this parametrization, in 1840 Rodrigues found a beautiful way of expressing the product of two rotations exp(ω1 n1 ) · exp(ω2 n2 ) in terms of the given data. In fact, this was an early exploration of the spin group! We refer to Whittaker [1988], Section 7, Altmann [1986] and references therein for further information. SU(2) Conjugacy Classes and the Hopf Fibration. We next determine all conjugacy classes of S 3 ∼ = SU(2). If a ∈ S 3 , then a−1 = a, and a straightforward computation gives aba−1 = (bs , 2(av · bv )av + 2as (av × bv ) + (2a2s − 1)bv )

5.2 Some Classical Lie Groups

325

for any b ∈ S 3 . If bs = ±1, that is, bv = 0, then the above formula shows that aba−1 = b for all a ∈ S 3 , that is, the classes of I and −I, where I = (1, 0), each consist of one element, and the center of SU(2) ∼ = S 3 is {±I}. In what follows, assume that bs 6= ±1, or, equivalently, that bv 6= 0, and fix this b ∈ S 3 throughout the following discussion. We shall prove that given x ∈ R3 with kxk = kbv k, we can find a ∈ S 3 such that 2(av · bv )av + 2as (av × bv ) + (2a2s − 1)bv = x.

(5.2.19)

If x = cbv for some c 6= 0, then the choice av = 0 and 2a2s = 1 + c satisfies (5.2.19). Now assume that x and bv are not collinear. Take the dot product of (5.2.19) with bv and get 2(av · bv )2 + 2a2s kbv k2 = kbv k2 + x · bv . If kbv k2 + x · bv = 0, since bv 6= 0, it follows that av · bv = 0 and as = 0. Returning to (5.2.19) it follows that −bv = x, which is excluded. Therefore, x · bv + kbv k2 6= 0, and searching for av ∈ R3 such that av · bv = 0, it follows that x · bv + kbv k2 6= 0. a2s = 2kbv k2 Now take the cross product of (5.2.19) with bv and recall that we assumed av · bv = 0 to get 2as kbv k2 av = bv × x, whence av =

bv × x , 2as kbv k2

which is allowed, since bv 6= 0 and as 6= 0. Note that a = (as , av ) just determined satisfies av · bv = 0 and |a|2 = a2s + kav k2 = 1, since kxk = kbv k. 5.2.21 Proposition.

∼ SU(2) are the two-spheres The conjugacy classes of S 3 =  bv ∈ R3 | kbv k2 = 1 − b2s

for each bs ∈ [−1, 1], which degenerate to the north and south poles (±1, 0, 0, 0) comprising the center of SU(2). The above proof shows that any unit quaternion is conjugate in S 3 to a quaternion of the form as + a3v k, as , a3v ∈ R, which in terms of matrices and the isomorphism (5.2.16) says that any SU(2) matrix is conjugate to a diagonal matrix . The conjugacy class of k is the unit sphere S 2 , and the orbit map π : S3 → S2,

π(a) = aka,

is the Hopf fibration. The subgroup H=



as + a3v k ∈ S 3 | as , a3v ∈ R



⊂ S3

is a closed, one-dimensional Abelian Lie subgroup of S 3 isomorphic via (5.2.16) to the set of diagonal matrices in SU(2) and is hence the circle S 1 . Note that the isotropy of k in S 3 consists of H, as an easy computation using (5.2.19) shows. Therefore, since the orbit of k is diffeomorphic to S 3 /H, it follows that the fibers of the Hopf fibration equal the left cosets aH for a ∈ S 3 .

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Finally, we shall give an expression of the Hopf fibration in terms of complex variables. In the representation (5.2.16), set w1 = x2 + ix1 , w2 = x0 + ix3 , and note that if a = (x0 , x1 , x2 , x3 ) ∈ S 3 ⊂ H, then aka corresponds to  0   0  x − ix3 −x2 − ix1 −i 0 x + ix3 x2 + ix1 x2 − ix1 x0 + ix3 0 i −x2 + ix1 x0 − ix3

  −i |x0 + ix3 |2 − |x2 + ix1 |2 −2i x2 + ix1 (x0 − ix3 )
. = −2i(x2 − ix1 )(x0 + ix3 ) i |x0 + ix3 |2 − |x2 + ix1 |2 Thus, if we consider the diffeomorphisms  −x2 − ix1 ∈ SU(2) x0 + ix3
7→ −i(x2 + ix1 ), −i(x0 + ix3 ) ∈ S 3 ⊂ C2 ,

(x0 , x1 , x2 , x3 ) ∈ S 3 ⊂ H 7→

 0 x − ix3 x2 − ix1

the above orbit map, that is, the Hopf fibration, becomes
(w1 , w2 ) ∈ S 3 7→ 2w1 w2 , |w2 |2 − |w1 |2 ∈ S 2 .

Exercises  5.2-1. Let SO(3) be the Lie group of 3 × 3 orthogonal matrices with determinant +1. Recall that the tangent space of SO(3) at the identity is given by the set of skew-symmetric matrices of the form   0 −ω3 ω2 0 −ω1 ,  ω ˆ = (ω)∧ =  ω3 ω = (ω1 , ω2 , ω3 ) ∈ R3 . −ω2 ω1 0 (i) Show that for R ∈ SO(3), the tangent space TR SO(3) consists of matrices of the form ω ˆ R where ω ˆ is skew-symmetric. (ii) Show that the flow of a vector field X(R) = ω ˆ R is given by φt (R) = exp(ˆ ω t)R where exp is the matrix exponential. (iii) Show that the Lie bracket between two vector fields X1 (R) = ω ˆ 1 R and X2 (R) = ω ˆ 2 R is given by [X1 , X2 ](R) = (ω 1 × ω 2 )∧ R, where × is the cross product in R3 . (iv) Describe the set of matrices in SO(3) that are also symmetric. (v) Show that the inner product on R3 and that on so(3) are related by ω1 · ω2 =


1 tr ω ˆ1ω ˆ 2T , 2

where ω1 , ω2 ∈ R3 .  5.2-2.

If A ∈ Sp(2n, R), show that AT ∈ Sp(2n, R) as well.

5.2 Some Classical Lie Groups

327

 5.2-3. Show that sp(2n, R) is isomorphic, as a Lie algebra, to the space of homogeneous quadratic functions on R2n under the Poisson bracket (see equation (5.2.15)).  5.2-4. A map f : Rn → Rn preserving the distance between any two points, that is, kf (x) − f (y)k = kx − yk for all x, y ∈ Rn , is called an isometry. Show that f is an isometry preserving the origin if and only if f ∈ O(n).  5.2-5.

The Euclidean Group.

(i) Let G = SO(n) × Rn and define a product on G by (A, v)(B, w) = (AB, Aw + v). Prove that G is a Lie group and identify the identity element and inverse for G. (G is the special Euclidean group, which is an example of a semi-direct product of Lie groups.) (ii) Show that SO(n) × {0} is a closed submanifold and subgroup of G.

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5. An Introduction to Lie Groups

Actions of Lie Groups

In this section we develop some basic facts about actions of Lie groups on manifolds. One of our main applications later will be the description of Hamiltonian systems with symmetry groups. Basic Definitions.

We begin with the definition of the action of a Lie group G on a manifold M .

5.3.1 Definition. Let M be a manifold and let G be a Lie group. A (left) action of a Lie group G on M is a smooth mapping Φ : G × M → M such that: (i) Φ(e, x) = x for all x ∈ M ; and (ii) Φ(g, Φ(h, x)) = Φ(gh, x) for all g, h ∈ G and x ∈ M . A right action is a map Ψ : M × G → M that satisfies Ψ(x, e) = x and Ψ(Ψ(x, g), h) = Ψ(x, gh). We sometimes use the notation g · x = Φ(g, x) for left actions, and x · g = Ψ(x, g) for right actions. In the infinite-dimensional case there are important situations where care with the smoothness is needed. For the formal development we assume that we are in the Banach–Lie group context. For every g ∈ G let Φg : M → M be given by x 7→ Φ(g, x). Then (i) becomes Φe = idM , while (ii) becomes Φgh = Φg ◦ Φh . Definition 5.3.1 can now be rephrased by saying that the map g 7→ Φg is a homomorphism of G into Diff(M ), the group of diffeomorphisms of M . In the special but important case where M is a Banach space V and each Φg : V → V is a continuous linear transformation, the action Φ of G on V is called a representation of G on V .

Examples (a) SO(3) acts on R3 by (A, x) 7→ Ax. This action leaves the two-sphere S 2 invariant, so the same formula defines an action of SO(3) on S 2 .  (b)

GL(n, R) acts on Rn by (A, x) 7→ Ax.



(c) Let X be a complete vector field on M , that is, one for which the flow Ft of X is defined for all t ∈ R. Then Ft : M → M defines an action of R on M .  Orbits and Isotropy.

If Φ is an action of G on M and x ∈ M , the orbit of x is defined by Orb(x) = { Φg (x) | g ∈ G } ⊂ M.

In finite dimensions one can show that Orb(x) is an immersed submanifold of M (Abraham and Marsden [1978, p. 265]). For x ∈ M , the isotropy (or stabilizer or symmetry ) group of Φ at x is given by Gx := { g ∈ G | Φg (x) = x } ⊂ G. Since the map Φx : G → M defined by Φx (g) = Φ(g, x) is continuous, Gx = (Φx )−1 (x) is a closed subgroup and hence a Lie subgroup of G. The manifold structure of Orb(x) is defined by requiring the bijective map [g] ∈ G/Gx 7→ g · x ∈ Orb(x) to be a diffeomorphism. That G/Gx is a smooth manifold follows from Proposition 9.3.2, which is discussed below. An action is said to be: 1. transitive if there is only one orbit or, equivalently, if for every x, y ∈ M there is a g ∈ G such that g · x = y; 2. effective (or faithful ) if Φg = idM implies g = e; that is, g 7→ Φg is one-to-one; and 3. free if it has no fixed points, that is, Φg (x) = x implies g = e or, equivalently, if for each x ∈ M , g 7→ Φg (x) is one-to-one. Note that an action is free iff Gx = {e}, for all x ∈ M and that every free action is faithful.

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329

Examples (a) Left translation. Lg : G → G, h 7→ gh, defines a transitive and free action of G on itself. Note that right multiplication Rg : G → G, h 7→ hg, does not define a left action because Rgh = Rh ◦ Rg , so that g 7→ Rg is an antihomomorphism. However, g 7→ Rg does define a right action, while g 7→ Rg−1 defines a left action of G on itself.  (b) Conjugation. G acts on G by conjugation, as follows: g 7→ Ig = Rg−1 ◦Lg . The map Ig : G → G given by h 7→ ghg −1 is the inner automorphism associated with g. Orbits of this action are called conjugacy classes or, in the case of matrix groups, similarity classes.  (c) Adjoint Action.

Differentiating conjugation at e, we get the adjoint representation of G on g: Adg := Te Ig : Te G = g → Te G = g.

Explicitly, the adjoint action of G on g is given by Ad : G × g → g,

Adg (ξ) = Te (Rg−1 ◦ Lg )ξ.

For example, for SO(3) we have IA (B) = ABA−1 , so differentiating with respect to B at B = identity gives ˆ = Aˆ AdA v vA−1 . However, ˆ )(w) = Aˆ (AdA v v(A−1 w) = A(v × A−1 w) = Av × w, so ˆ ) = (Av)ˆ. (AdA v Identifying so(3) ∼ = R3 , we get AdA v = Av.



(d) Coadjoint Action. The coadjoint action of G on g∗ , the dual of the Lie algebra g of G, is defined as follows. Let Ad∗g : g∗ → g∗ be the dual of Adg , defined by

∗  Adg α, ξ = hα, Adg ξi for α ∈ g∗ and ξ ∈ g. Then the map Φ∗ : G × g∗ → g∗

given by

(g, α) 7→ Ad∗g−1 α

is the coadjoint action of G on g∗ . The corresponding coadjoint representation of G on g∗ is denoted by
∗ Ad∗ : G → GL(g∗ , g∗ ), Ad∗g−1 = Te (Rg ◦ Lg−1 ) . We will avoid the introduction of yet another ∗ by writing (Adg−1 )∗ or simply Ad∗g−1 , where ∗ denotes the usual linear-algebraic dual, rather than Ad∗ (g), in which ∗ is simply part of the name of the function Ad∗ . Any representation of G on a vector space V similarly induces a contragredient representation of G on V ∗.  Quotient (Orbit) Spaces. An action of Φ of G on a manifold M defines an equivalence relation on M by the relation of belonging to the same orbit; explicitly, for x, y ∈ M , we write x ∼ y if there exists a g ∈ G such that g · x = y, that is, if y ∈ Orb(x) (and hence x ∈ Orb(y)). We let M/G be the set of these equivalence classes, that is, the set of orbits, sometimes called the orbit space. Let π : M → M/G,

x 7→ Orb(x),

and give M/G the quotient topology by defining U ⊂ M/G to be open if and only if π −1 (U ) is open in M . To guarantee that the orbit space M/G has a smooth manifold structure, further conditions on the action are required.

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An action Φ : G × M → M is called proper if the mapping ˜ : G × M → M × M, Φ defined by ˜ x) = (x, Φ(g, x)), Φ(g, ˜ −1 (K) is compact. In is proper. In finite dimensions this means that if K ⊂ M × M is compact, then Φ general, this means that if {xn } is a convergent sequence in M and {Φgn (xn )} converges in M , then {gn } has a convergent subsequence in G. For instance, if G is compact, this condition is automatically satisfied. Orbits of proper Lie group actions are closed and hence embedded submanifolds. The next proposition gives a useful sufficient condition for M/G to be a smooth manifold. 5.3.2 Proposition. If Φ : G × M → M is a proper and free action, then M/G is a smooth manifold and π : M → M/G is a smooth submersion. For the proof, one uses the results in Chapter 3 on quotient manifolds; see also Proposition 4.2.23 in Abraham and Marsden [1978]. (In infinite dimensions one uses these ideas, but additional technicalities often arise; see Ebin [1970] and Isenberg and Marsden [1982].) The idea of the chart construction for M/G is based on the following observation. If x ∈ M , then there is an isomorphism ϕx of Tπ(x) (M/G) with the quotient space Tx M/Tx Orb(x). Moreover, if y = Φg (x), then Tx Φg induces an isomorphism ψx,y : Tx M/Tx Orb(x) → Ty M/Ty Orb(y) satisfying ϕy ◦ ψx,y = ϕx .

Examples (a)

G = R acts on M = R by translations; explicitly, Φ : G × M → M,

Φ(s, x) = x + s.

Then for x ∈ R, Orb(x) = R. Hence M/G is a single point, and the action is transitive, proper, and free.  (b) G = SO(3), M = R3 (∼ = so(3)∗ ). Consider the action for x ∈ R3 and A ∈ SO(3) given by ΦA x = Ax. Then Orb(x) = { y ∈ R3 | kyk = kxk } = a sphere of radius kxk. Hence M/G ∼ = R+ . The set R+ = { r ∈ R | r ≥ 0 } is not a manifold because it includes the endpoint r = 0. Indeed, the action is not free, since it has the fixed point 0 ∈ R3 .  (c) Let G be Abelian. Then Adg = idg , Ad∗g−1 = idg∗ , and the adjoint and coadjoint orbits of ξ ∈ g and α ∈ g∗ , respectively, are the one-point sets {ξ} and {α}.  We will see later that coadjoint orbits can be natural phase spaces for some mechanical systems like the rigid body; in particular, they are always even-dimensional.

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Infinitesimal Generators. Next we turn to the infinitesimal description of an action, which will be a crucial concept for mechanics. 5.3.3 Definition. by

Suppose Φ : G × M → M is an action. For ξ ∈ g, the map Φξ : R × M → M , defined Φξ (t, x) = Φ(exp tξ, x),

is an R-action on M . In other words, Φexp tξ : M → M is a flow on M . The corresponding vector field on M , given by d Φexp tξ (x), ξM (x) := dt t=0 is called the infinitesimal generator of the action corresponding to ξ. 5.3.4 Proposition.

The tangent space at x to an orbit Orb(x0 ) is Tx Orb(x0 ) = { ξM (x) | ξ ∈ g } ,

where Orb(x0 ) is endowed with the manifold structure making G/Gx0 → Orb(x0 ) into a diffeomorphism. The idea is as follows: Let σξ (t) be a curve in G with σξ (0) = e that is tangent to ξ at t = 0. Then the map Φx,ξ (t) = Φσξ (t) (x) is a smooth curve in Orb(x0 ) with Φx,ξ (0) = x. Hence by the chain rule (see also Lemma 9.3.7 below), d d x,ξ Φ (t) = Φσ (t) (x) = ξM (x) dt t=0 dt t=0 ξ is a tangent vector at x to Orb(x0 ). Furthermore, each tangent vector is obtained in this way, since tangent vectors are equivalence classes of such curves. The Lie algebra of the isotropy group Gx , x ∈ M , called the isotropy (or stabilizer, or symmetry) algebra at x, equals, by Proposition 5.1.13, gx = { ξ ∈ g | ξM (x) = 0 }.

Examples (a)

The infinitesimal generators for the adjoint action are computed as follows. Let Ad : G × g → g,

Adg (η) = Te (Rg−1 ◦ Lg )(η).

For ξ ∈ g, we compute the corresponding infinitesimal generator ξg . By definition,   d ξg (η) = Adexp tξ (η). dt t=0 By (5.1.5), this equals [ξ, η]. Thus, for the adjoint action, ξg (η) = [ξ, η].

(5.3.1)

This important operation deserves a special name. We define the ad operator adξ : g → g by η 7→ [ξ, η]. Thus, ξg = adξ .  (b) We illustrate (a) for the group SO(3) as follows. Let A(t) = exp(tC), where C ∈ so(3); then A(0) = I and A0 (0) = C. Thus, with B ∈ so(3), d d (Adexp tC B) = (exp(tC)B(exp(tC))−1 ) dt t=0 dt t=0 d = (A(t)BA(t)−1 ) dt t=0 = A0 (0)BA−1 (0) + A(0)BA−10 (0).

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Differentiating A(t)A−1 (t) = I, we obtain d −1 (A (t)) = −A−1 (t)A0 (t)A−1 (t), dt so that A−10 (0) = −A0 (0) = −C. Then the preceding equation becomes d (Adexp tC B) = CB − BC = [C, B], dt t=0 as expected.



(c) Let Ad∗ : G × g∗ → g∗ be the coadjoint action (g, α) 7→ Ad∗g−1 α. If ξ ∈ g, we compute for α ∈ g∗ and η∈g   d ∗ hξg∗ (α), ηi = (α), η Ad exp(−tξ) dt t=0 D E

 d d ∗ = α, Adexp(−tξ) η Ad (α), η = exp(−tξ) dt dt t=0   t=0 d Adexp(−tξ) η = α, dt t=0 

= hα, −[ξ, η]i = − hα, adξ (η)i = − ad∗ξ (α), η . Hence ξg∗ = − ad∗ξ ,

or ξg∗ (α) = − hα, [ξ, ·]i .

(5.3.2) 

∗

(d) Identifying so(3) ∼ = (R3 , ×) and so(3)∗ ∼ = R3 , using the pairing given by the standard Euclidean inner product, (5.3.2) reads ξso(3)∗ (l) = −l · (ξ × ·), for l ∈ so(3)∗ and ξ ∈ so(3). For eprint = ....., η ∈ so(3), we have 

ξso(3)∗ (l), η = −l · (ξ × η) = −(l × ξ) · η = −hl × ξ, ηi, so that ξR3 (l) = −l × ξ = ξ × l. As expected, ξR3 (l) ∈ Tl Orb(l) is tangent to Orb(l) (see Figure 9.3.1). Allowing ξ to vary in so(3) ∼ = R3 , one obtains all of Tl Orb(l), consistent with Proposition 9.3.4.  Equivariance. A map between two spaces is equivariant when it respects group actions on these spaces. We state this more precisely: 5.3.5 Definition. Let M and N be manifolds and let G be a Lie group that acts on M by Φg : M → M , and on N by Ψg : N → N . A smooth map f : M → N is called equivariant with respect to these actions if for all g ∈ G, f ◦ Φg = Ψg ◦ f, (5.3.3) that is, if the diagram in Figure 9.3.2 commutes. Setting g = exp(tξ) and differentiating (5.3.3) with respect to t at t = 0 gives T f ◦ ξM = ξN ◦ f . In other words, ξM and ξN are f -related. In particular, if f is an equivariant diffeomorphism, then f ∗ ξN = ξM . Also note that if M/G and N/G are both smooth manifolds with the canonical projections smooth submersions, an equivariant map f : M → N induces a smooth map fG : M/G → N/G.

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l

Figure 5.3.1. ξR3 (l) is tangent to Orb(l).

M

f -

N

-

Ψg ? N

Φg ? M f

Figure 5.3.2. Commutative diagram for equivariance.

Averaging. A useful device for constructing invariant objects is by averaging. For example, let G be a compact group acting on a manifold M and let α be a differential form on M . Then we form Z α= Φ∗g α dµ(g), G

where µ is Haar measure on G. One checks that α is invariant. One can do the same with other tensors, such as Riemannian metrics on M , to obtain invariant ones. Brackets of Generators. Now we come to an important formula relating the Jacobi–Lie bracket of two infinitesimal generators with the Lie algebra bracket. 5.3.6 Proposition. Let the Lie group G act on the left on the manifold M . Then the infinitesimal generator map ξ 7→ ξM of the Lie algebra g of G into the Lie algebra X(M ) of vector fields of M is a Lie algebra antihomomorphism; that is, (aξ + bη)M = aξM + bηM and [ξM , ηM ] = −[ξ, η]M for all ξ, η ∈ g and a, b ∈ R. To prove this, we use the following lemma: 5.3.7 Lemma.

(i) Let c(t) be a curve in G, c(0) = e, c0 (0) = ξ ∈ g. Then d ξM (x) = Φc(t) (x). dt t=0

(ii) For every g ∈ G, (Adg ξ)M = Φ∗g−1 ξM .

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(i) Let Φx : G → M be the map Φx (g) = Φ(g, x). Since Φx is smooth, the definition of the Proof. infinitesimal generator says that Te Φx (ξ) = ξM (x). Thus, (i) follows by the chain rule. (ii) We have d Φ(exp(t Adg ξ), x) dt t=0 d Φ(g(exp tξ)g −1 , x) (by Corollary 9.1.8) = dt t=0 d = (Φg ◦ Φexp tξ ◦ Φg−1 (x)) dt t=0

= TΦ−1 Φg ξM Φg−1 (x) g (x)   = Φ∗g−1 ξM (x).

(Adg ξ)M (x) =



Proof of Proposition 9.3.6. Linearity follows, since ξM (x) = Te Φx (ξ). To prove the second relation, put g = exp tη in (ii) of the lemma to get (Adexp tη ξ)M = Φ∗exp(−tη) ξM . But Φexp(−tη) is the flow of −ηM , so differentiating at t = 0 the right-hand side gives [ξM , ηM ]. The derivative of the left-hand side at t = 0 equals [η, ξ]M by the preceding Example (a).  In view of this proposition one defines a left Lie algebra action of a manifold M as a Lie algebra antihomomorphism ξ ∈ g 7→ ξM ∈ X(M ), such that the mapping (ξ, x) ∈ g × M 7→ ξM (x) ∈ T M is smooth. Let Φ : G × G → G denote the action of G on itself by left translation: Φ(g, h) = Lg h. For ξ ∈ g, let Yξ be the corresponding right-invariant vector field on G. Then ξG (g) = Yξ (g) = Te Rg (ξ), and similarly, the infinitesimal generator of right translation is the left-invariant vector field g 7→ Te Lg (ξ). Derivatives of Curves. It is convenient to have formulas for the derivatives of curves associated with the adjoint and coadjoint actions. For example, let g(t) be a (smooth) curve in G and η(t) a (smooth) curve in g. Let the action be denoted by concatenation: g(t)η(t) = Adg(t) η(t). 5.3.8 Proposition.

The following holds:   dη d g(t)η(t) = g(t) [ξ(t), η(t)] + , dt dt

where ξ(t) = g(t)−1 g(t) ˙ := Tg(t) L−1 g(t) Proof.

dg ∈ g. dt

We have  d d Adg(t) η(t) = g(t0 )[g(t0 )−1 g(t)]η(t) dt t=t0 dt t=t0  d [g(t0 )−1 g(t)]η(t) , = g(t0 ) dt t=t0

(5.3.4)

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where the first g(t0 ) denotes the Ad-action, which is linear . Now, g(t0 )−1 g(t) is a curve through the identity at t = t0 with tangent vector ξ(t0 ), so the above becomes   dη(t0 ) . g(t0 ) [ξ(t0 ), η(t0 )] + dt  Similarly, for the coadjoint action we write g(t)µ(t) = Ad∗g(t)−1 µ(t), and then, as above, one proves that   d dµ ∗ [g(t)µ(t)] = g(t) − adξ(t) µ(t) + , dt dt which we could write, extending our concatenation notation to Lie algebra actions as well,   dµ d [g(t)µ(t)] = g(t) ξ(t)µ(t) + , dt dt where ξ(t) = g(t)−1 g(t). ˙ For right actions, these become   d dη [η(t)g(t)] = η(t)ζ(t) + g(t) dt dt and

d [µ(t)g(t)] = dt

  dµ µ(t)ζ(t) + g(t), dt

(5.3.5)

(5.3.6)

(5.3.7)

−1 where ζ(t) = g(t)g(t) ˙ ,

η(t)g(t) = Adg(t)−1 η(t),

and η(t)ζ(t) = −[ζ(t), η(t)],

and where µ(t)g(t) = Ad∗g(t) µ(t) Connectivity of Some Classical Groups.

and µ(t)ζ(t) = ad∗ζ(t) µ(t).

First we state two facts about homogeneous spaces:

1. If H is a closed normal subgroup of the Lie group G (that is, if h ∈ H and g ∈ G, then ghg −1 ∈ H), then the quotient G/H is a Lie group and the natural projection π : G → G/H is a smooth group homomorphism. (This follows from Proposition 9.3.2; see also Theorem 2.9.6 in Varadarajan [1974, p. 80].) Moreover, if H and G/H are connected, then G is connected. Similarly, if H and G/H are simply connected, then G is simply connected. 2. Let G, M be finite-dimensional and second countable and let Φ : G × M → M be a transitive action of G on M , and for x ∈ M , let Gx be the isotropy subgroup of x. Then the map gGx 7→ Φg (x) is a diffeomorphism of G/Gx onto M . (This follows from Proposition 9.3.2; see also Theorem 2.9.4 in Varadarajan [1974, p. 77].) The action Φ : GL(n, R) × Rn → Rn ,

Φ(A, x) = Ax,

restricted to O(n) × S n−1 induces a transitive action. The isotropy subgroup of O(n) at en ∈ S n−1 is O(n − 1). Clearly, O(n − 1) is a closed subgroup of O(n) by embedding any A ∈ O(n − 1) as   A 0 A˜ = ∈ O(n), 0 1

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and the elements of O(n−1) leave en fixed. On the other hand, if A ∈ O(n) and Aen = en , then A ∈ O(n−1). It follows from fact 2 above that the map O(n)/ O(n − 1) → S n−1 ,

A · O(n − 1) 7→ Aen ,

is a diffeomorphism. By a similar argument, there is a diffeomorphism S n−1 ∼ = SO(n)/ SO(n − 1). The natural action of GL(n, C) on Cn similarly induces a diffeomorphism of S 2n−1 ⊂ R2n with the homogeneous space U(n)/ U(n − 1). Moreover, we get S 2n−1 ∼ = SU(n)/ SU(n − 1). In particular, since SU(1) consists only of the 1 × 1 identity matrix, S 3 is diffeomorphic with SU(2), a fact already proved at the end of §9.2. 5.3.9 Proposition. Each of the Lie groups SO(n), SU(n), and U(n) is connected for n ≥ 1, and O(n) has two components. The group SU(n) is simply connected. Proof. The groups SO(1) and SU(1) are connected, since both consist only of the 1 × 1 identity matrix, and U(1) is connected, since U(1) = { z ∈ C | |z| = 1 } = S 1 . That SO(n), SU(n), and U(n) are connected for all n now follows from fact 1 above, using induction on n and the representation of the spheres as homogeneous spaces. Since every matrix A in O(n) has determinant ±1, the orthogonal group can be written as the union of two nonempty disjoint connected open subsets as follows: O(n) = SO(n) ∪ A · SO(n), where A = diag(−1, 1, 1, . . . , 1). Thus, O(n) has two components.



Here is a general strategy for proving the connectivity of the classical groups; see, for example Knapp [2002]. This works, in particular, for Sp(2n, R) (and the groups Sp(2n, C), SP∗ (2n) discussed in the Internet supplement). Let G be a subgroup of GL(n, R) (resp. GL(n, C)) defined as the zero set of a collection of real-valued polynomials in the (real and imaginary parts) of the matrix entries. Assume also that G is closed under taking adjoints (see Exercise 9.2-2 for the case of Sp(2n, R)). Let K = G ∩ O(n) (resp. U(n)) and let p be the set of Hermitian matrices in g. The polar decomposition says that (k, ξ) ∈ K × p 7→ k exp(ξ) ∈ G is a homeomorphism. It follows that since ξ lies in a connected space, G is connected iff K is connected. For Sp(2m, R) our results above show that U(m) is connected, so Sp(2m, R) is connected.

Examples (a) Isometry groups. Let E be a finite-dimensional vector space with a bilinear form h , i. Let G be the group of isometries of E, that is, F is an isomorphism of E onto E and hF e, F e0 i = he, e0 i, for all e and e0 ∈ E. Then G is a subgroup and a closed submanifold of GL(E). The Lie algebra of G is { K ∈ L(E) | hKe, e0 i + he, Ke0 i = 0 for all e, e0 ∈ E }.

(b) Lorentz group.

If h , i denotes the Minkowski metric on R4 , that is, hx, yi =

3 X i=1

xi y i − x4 y 4 ,



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then the group of linear isometries is called the Lorentz group L. The dimension of L is six, and L has four connected components. If   I3 0 S= ∈ GL(4, R), 0 −1 then L = { A ∈ GL(4, R) | AT SA = S }, and so the Lie algebra of L is l = { A ∈ L(R4 , R4 ) | SA + AT S = 0 }. The identity component of L is { A ∈ L | det A > 0 and A44 > 0 } = L+ ↑; L and L+ ↑ are not compact.



(c) Galilean group. Consider the (closed) subgroup G following block structure:  R {R, v, a, τ } :=  0 0

of GL(5, R) that consists of matrices with the v 1 0

 a τ , 1

where R ∈ SO(3), v, a ∈ R3 , and τ ∈ R. This group is called the Galilean group. Its Lie algebra is a subalgebra of L(R5 , R5 ) given by the set of matrices of the form   ˆ u α ω {ω, u, α, θ} :=  0 0 θ  , 0 0 0 where ω, u, α ∈ R3 and θ ∈ R. Obviously the Galilean group acts naturally on R5 ; moreover, it acts naturally on R4 , embedded as the following G-invariant subset of R5 :     x x 7→  t  , t 1 where x ∈ R3 and t ∈ R. Concretely, the action of {R, v, a, τ } on (x, t) is given by (x, t) 7→ (Rx + tv + a, t + τ ). Thus, the Galilean group gives a change of frame of reference (not affecting the “absolute time” variable) by rotations (R), space translations (a), time translations (τ ), and going to a moving frame, or boosts (v).  (d) Unitary Group of Hilbert Space. Another basic example of an infinite-dimensional group is the unitary group U(H) of a complex Hilbert space H. If G is a Lie group and ρ : G → U(H) is a group homomorphism, we call ρ a unitary representation. In other words, ρ is an action of G on H by unitary maps. As with the diffeomorphism group, questions of smoothness regarding U(H) need to be dealt with carefully, and in this book we shall give only a brief indication of what is involved. The reason for care is, for one thing, that one ultimately is dealing with PDEs rather than ODEs and the hypotheses made must be such that PDEs are not excluded. For example, for a unitary representation one assumes that for each ψ, ϕ ∈ H,

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the map g 7→ hψ, ρ(g)ϕi of G to C is continuous. In particular, for G = R, one has the notion of a continuous one-parameter group U (t) of unitary operators4 so that U (0) = identity and U (t + s) = U (t) ◦ U (s). Stone’s theorem says that in an appropriate sense we can write U (t) = etA , where A is an (unbounded) skew-adjoint operator defined on a dense domain D(A) ⊂ H. See Section 7.4B for the proof. Conversely each skew-adjoint operator defines a one-parameter subgroup. Thus, Stone’s theorem gives precise meaning to the statement that the Lie algebra u(H) of U(H) consists of the skew-adjoint operators. The Lie bracket is the commutator, as long as one is careful with domains. If ρ is a unitary representation of a finite-dimensional Lie group G on H, then ρ(exp(tξ)) is a one-parameter subgroup of U(H), so Stone’s theorem guarantees that there is a map ξ 7→ A(ξ) associating a skew-adjoint operator A(ξ) to each ξ ∈ g. Formally, we have [A(ξ), A(η)] = A[ξ, η]. Results like this are aided by a theorem of Nelson [1959] guaranteeing a dense subspace DG ⊂ H such that (i) A(ξ) is well-defined on DG , (ii) A(ξ) maps DG to DG , and (iii) for ψ ∈ DG , [exp tA(ξ)]ψ is C ∞ in t with derivative at t = 0 given by A(ξ)ψ. This space is called an essential G-smooth part of H, and on DG the above commutator relation and the linearity A(αξ + βη) = αA(ξ) + βA(η) become literally true. Moreover, we lose little by using DG , since A(ξ) is uniquely determined by what it is on DG . We identify U(1) with the unit circle in C, and each such complex number determines an element of U(H) by multiplication. Thus, we regard U(1) ⊂ U(H). As such, it is a normal subgroup (in fact, elements of U(1) commute with elements of U(H)), so the quotient is a group, called the projective unitary group of H. We write it as U(PH) = U(H)/ U(1). We write elements of U(PH) as [U ] regarded as an equivalence class of U ∈ U(H). The group U(PH) acts on projective Hilbert space PH = H/C, as in §5.3, by [U ][ϕ] = [U ϕ]. One-parameter subgroups of U(PH) are of the form [U (t)] for a one-parameter subgroup U (t) of U(H). This is a particularly simple case of the general problem considered by Bargmann and Wigner of lifting projective representations, a topic we return to later. In any case, this means that we can identify the Lie algebra as u(PH) = u(H)/iR, where we identify the two skew-adjoint operators A and A + λi, for λ real. A projective representation of a group G is a homomorphism τ : G → U(PH); we require continuity of g ∈ G 7→ |hψ, τ (g)ϕi| ∈ R, which is well-defined for [ψ], [ϕ] ∈ PH. There is an analogue of Nelson’s theorem that guarantees an essential G-smooth part PDG of PH with properties like those of DG .  Miscellany.

We conclude this section with a variety of remarks.

1. Coadjoint Isotropy. The first remark concerns coadjoint orbit isotropy groups. The main result here is the following theorem, due to Duflo and Vergne [1969]. (An alternative proof may be found in Rais [1972]). 5.3.10 Theorem (Duflo and Vergne). Letg be a finite-dimensional Lie algebra with dual g∗ and let r = min { dim gµ | µ ∈ g∗ }. The set { µ ∈ g∗ | dim gµ = r } is open and dense in g∗ . If dim gµ = r, then gµ is Abelian. A simple example is the rotation group SO(3) in which the coadjoint isotropy at each nonzero point is the Abelian group S 1 , whereas at the origin it is the nonabelian group SO(3). 4 In Hilbert space, weak continuity and unitarity implies continuity in the operator norm; see, for example, Riesz and Sz.-Nagy [1952] §29.

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2. Remarks on Infinite-Dimensional Groups. We can use a slight reinterpretation of the formulae in this section to calculate the Lie algebra structure of some infinite-dimensional groups. Here we will treat this topic only formally, that is, we assume that the spaces involved are manifolds and do not specify the function-space topologies. For the formal calculations, these structures are not needed, but the reader should be aware that there is a mathematical gap here. (See Ebin and Marsden [1970] and Adams, Ratiu, and Schmid [1986]) for more information.) Given a manifold M , let Diff(M ) denote the group of all diffeomorphisms of M . The group operation is composition. The Lie algebra of Diff(M ), as a vector space, consists of vector fields on M ; indeed, the flow of a vector field is a curve in Diff(M ), and its tangent vector at t = 0 is the given vector field. To determine the Lie algebra bracket, we consider the action of an arbitrary Lie group G on M . Such an action of G on M may be regarded as a homomorphism Φ : G → Diff(M ). By Proposition 9.1.5, its derivative at the identity Te Φ should be a Lie algebra homomorphism. From the definition of infinitesimal generator, we see that Te Φ · ξ = ξM . Thus, Proposition 9.1.5 suggests that [ξM , ηM ]Lie bracket = [ξ, η]M . However, by Proposition 9.3.6, [ξ, η]M = −[ξM , ηM ]. Thus, [ξM , ηM ]Lie bracket = −[ξM , ηM ]. This suggests that the Lie algebra bracket on X(M ) is minus the Jacobi–Lie bracket. Another way to arrive at the same conclusion is to use the method of computing brackets in the table in §9.1. To do this, we first compute, according to step 1, the inner automorphism to be Iη (ϕ) = η ◦ ϕ ◦ η −1 . By step 2, we differentiate with respect to ϕ to compute the Ad map. Letting X be the time derivative at t = 0 of a curve ϕt in Diff(M ) with ϕ0 = Identity, we have 

 d d ϕt = Iη (ϕt ) dt t=0 dt t=0

Adη (X) = (Te Iη )(X) = Te Iη d = (η ◦ ϕt ◦ η −1 ) = T η ◦ X ◦ η −1 = η∗ X. dt t=0

Hence Adη (X) = η∗ X. Thus, the adjoint action of Diff(M ) on its Lie algebra is just the push-forward operation on vector fields. Finally, as in step 3, we compute the bracket by differentiating Adη (X) with respect to η. But by the Lie derivative characterization of brackets and the fact that push-forward is the inverse of pull-back, we arrive at the same conclusion. In summary, either method suggests that The Lie algebra bracket on Diff(M ) is minus the Jacobi–Lie bracket of vector fields. One can also say that the Jacobi–Lie bracket gives the right (as opposed to left) Lie algebra structure on Diff(M ). If one restricts to the group of volume-preserving (or symplectic) diffeomorphisms, then the Lie bracket is again minus the Jacobi–Lie bracket on the space of divergence-free (or locally Hamiltonian) vector fields. Here are three examples of actions of Diff(M ). Firstly, Diff(M ) acts on M by evaluation: The action Φ : Diff(M ) × M → M is given by Φ(ϕ, x) = ϕ(x). Secondly, the calculations we did for Adη show that the adjoint action of Diff(M ) on its Lie algebra is given by push-forward. Thirdly, if we identify the dual space X(M )∗ with one-form densities by means of integration, then the change-of-variables formula shows that the coadjoint action is given by push-forward of one-form densities.

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Exercises  5.3-1.

Let a Lie group G act linearly on a vector space V . Define a group structure on G × V by (g1 , v1 ) · (g2 , v2 ) = (g1 g2 , g1 v2 + v1 ).

Show that this makes G×V into a Lie group—it is called the semidirect product and is denoted by G s V . Determine its Lie algebra g s V .  5.3-2. (a) Show that the Euclidean group E(3) can be written as O(3) s R3 in the sense of the preceding exercise (see also Exercise 5.2-5). (b) Show that E(3) is isomorphic to the group of 4 × 4 matrices of the form   A b , 0 1 where A ∈ O(3) and b ∈ R3 .  5.3-3. Show that the Galilean group may be written as a semidirect product G = (SO(3) s R3 ) s R4 . Compute explicitly the inverse of a group element, and the adjoint and the coadjoint actions.  5.3-4. If G is a Lie group, show that T G is isomorphic (as a Lie group) with G s g (see Exercises 5.1-2 and 5.3-1).  5.3-5. orbits.

Compute the coadjoint action of SO(3) on the dual of its Lie algebra and also find the coadjoint

 5.3-6.

Verify, using standard vector notation, the four “derivative of curves” formulas for SO(3).

 5.3-7. Use the complex polar decomposition theorem (Proposition 5.2.15) and simple connectedness of SU(n) to show that SL(n, C) is also simply connected.  5.3-8. Show that SL(2, C) is the simply connected covering group of the identity component L†↑ of the Lorentz group.

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6 Tensors

This version: February 4, 2007

In the previous chapter we studied vector fields and functions on manifolds. In this chapter these objects are generalized to tensor fields, which are sections of vector bundles built out of the tangent bundle. This study is continued in the next chapter when we discuss differential forms, which are tensors with special symmetry properties. One of the objectives of this chapter is to extend the pull-back and Lie derivative operations from functions and vector fields to tensor fields.

6.1

Tensors on Linear Spaces

Preparatory to putting tensors on manifolds, we first study them on vector spaces. This subject is an extension of linear algebra sometimes called “multilinear algebra.” Ultimately our constructions will be done on each fiber of the tangent bundle, producing a new vector bundle. Tensors and their Components. As in Chapter 2, E, F, . . . denote Banach spaces and Lk (E1 , . . . , Ek ; F) denotes the vector space of continuous k-multilinear maps of E1 × · · · × Ek to F. The special case L(E, R) is denoted E∗ , the dual space of E. If E is finite dimensional and {e1 , . . . , en } isan ordered basis of E, there is a unique ordered basis of E∗ , the dual basis {e1 , . . . , en }, such that ej , ei = δij where δij = 1 if j = i and 0 otherwise. Here, h , i denotes the natural pairing between E and E∗ . Furthermore, for each v ∈ E, v=

n X

i=1

 ei , v ei

and α =

n X

hα, ei i ei ,

i=1

∗

for each α ∈ E . Employing the summation convention whereby summation is implied when an index is repeated on upper and lower levels, these expressions become

 v = ei , v ei and α = hα, ei i ei . As in Supplement 2.4C, if E is infinite dimensional, by E∗ we will mean another Banach space weakly paired to E; it need not be the full functional analytic dual of E. In particular, E∗∗ will always be chosen to be E. With these conventions, tensors are defined as follows.

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6.1.1 Definition.

For a vector space E we put Tsr (E) = Lr+s (E∗ , . . . , E∗ , E, . . . , E; R)

(r copies of E∗ and s copies of E). Elements of Tsr (E) are called tensors on E, contravariant of order r and covariant of order s; or simply, of type (r, s). +r2 (E) Given t1 ∈ Tsr11 (E) and t2 ∈ Tsr22 (E), the tensor product of t1 and t2 is the tensor t1 ⊗ t2 ∈ Tsr11+s 2 defined by (t1 ⊗ t2 )(β 1 , . . . , β r1 , γ 1 , . . . , γ r2 , f1 , . . . , fs1 , g1 , . . . , gs2 ) = t1 (β 1 , . . . , β r1 , f1 , . . . , fs1 )t2 (γ 1 , . . . , γ r2 , g1 , . . . , gs2 ) where β j , γ j ∈ E∗ and fj , gj ∈ E. Replacing R by a space F gives Tsr (E; F), the F-valued tensors of type (r, s). The tensor product now requires a bilinear form on the value space for its definition. For R-valued tensors, ⊗ is associative, bilinear and continuous; it is not commutative. We also have the special cases T01 (E) = E, T10 (E) = E∗ , T20 (E) = L(E; E∗ ), and T11 (E) = L(E; E) and make the convention that T00 (E; F) = F. 6.1.2 Proposition. Let E be an n dimensional vector space. If {e1 , . . . , en } is a basis of E and {e1 , . . . , en } is the dual basis, then { ei1 ⊗ · · · ⊗ eir ⊗ ej1 ⊗ · · · ⊗ ejs | i1 , . . . , ir , j1 , . . . , js = 1, . . . , n } is a basis of Tsr (E) and thus dim(Tsr (E)) = nr+s . Proof.

We must show that the elements ei1 ⊗ · · · ⊗ eir ⊗ ej1 ⊗ · · · ⊗ ejs

of Tsr (E) are linearly independent and span Tsr (E). Suppose that a finite sum vanishes: j1 js r tij11...i ...js ei1 ⊗ · · · ⊗ eir ⊗ e ⊗ · · · ⊗ e = 0. ...kr = 0. Next, check that for t ∈ Tsr (E) we have Apply this to (ek1 , . . . , ekr , e`1 , . . . , e`s ) to get tk`11...` s

t = t(ei1 , . . . , eir , ej1 , . . . , ejs )ei1 ⊗ · · · ⊗ eir ⊗ ej1 ⊗ · · · ⊗ ejs .



The coefficients i1 ir r tij11...i ...js = t(e , . . . , e , ej1 , . . . , ejs )

are called the components of t relative to the basis {e1 , . . . , en }. 6.1.3 Examples. A. If t is a (0, 2)-tensor on E then t has components tij = t(ei , ej ), an n × n matrix. This is the usual way of associating a bilinear form with a matrix. For instance, in R2 the bilinear form t(x, y) = Ax1 y1 + Bx1 y2 + Cx2 y1 + Dx2 y2 (where x = (x1 , x2 ) and y = (y1 , y2 )) is associated to the 2 × 2 matrix   A B . C D

6.1 Tensors on Linear Spaces

343

B. If t is a (0, 2)-tensor on E, it makes sense to say that t is symmetric; that is, t(ei , ej ) = t(ej , ei ). This is equivalent to saying that the matrix [tij ] is symmetric. Symmetric (0, 2)–tensors t can be recovered from their quadratic form Q(e) = t(e, e)

by t (e1 , e2 ) =

1 [Q (e1 + e2 ) − Q (e1 − e2 )] , 4

the polarization identity . If E = R2 and t has the matrix   A B . B C then Q(x) = Ax21 + 2Bx1 x2 + Cx22 . Symmetric (0, 2)-tensors are thus closely related to quadratic forms and arise, for example, in mechanics as moment of inertia tensors and stress tensors. C.

In general, a symmetric (r, 0)-tensor is defined by the condition t(α1 , . . . , αr ) = t(ασ(1) , . . . , ασ(r) )

for all permutations σ of {1, . . . , r}, and all elements α1 , . . . , αr ∈ E∗ . One may associate to t a homogeneous polynomial of degree r, P (α) = t(α, . . . , α) and as in the case r = 2, P and t determine each other. A similar definition holds for (0, s)-tensors. It is clear that a tensor is symmetric iff all its components in an arbitrary basis are symmetric. D. An inner product h , i on E is a symmetric (0, 2)-tensor. Its matrix has components gij = hei , ej i. Thus gij is symmetric and positive definite. The components of the inverse matrix are written g ij . E. The space Lk (E1 , . . . , Ek ; F) is isometric to Lk (Eσ(1) , . . . , Eσ(k) ; F) for any permutation σ of {1, . . . , k}, the isometry being given by A 7→ A0 , where A0 (eσ(1) , . . . , eσ(k) ) = A(e1 , . . . , ek ). Thus if t ∈ Tsr (E; F), the tensor t can be regarded in C(r + s, s) (the number of ways r + s objects chosen s at a time) ways as an (r + s)-multilinear F-valued map. For example, if t ∈ T12 (E), the standard way is to regard it as a 3-linear map t : E∗ × E∗ × E → R. There are two more ways to interpret this map, however, namely as E∗ × E × E∗ → R and as E × E∗ × E∗ → R. In finite dimensions, where one writes the tensors in components, this distinction is important and is reflected in the index positions. Thus the three different tensors described above are written k tij k ei ⊗ ej ⊗ e ,

k tij k ei ⊗ e ⊗ ej ,

k tij k e ⊗ ei ⊗ ej .

F. In classical mechanics one encounters the notion of a dyadic (cf. Goldstein [1980]). A dyadic is the formal sum of a finite number of dyads, a dyad being a pair of vectors e1 , e2 ∈ R3 written in a specific order in the form e1 e2 . The action of a dyad on a pair of vectors, called the double dot product of two dyads is defined by e1 e2 : u1 u2 = (e1 · u1 )(e2 · u2 ), where · stands for the usual dot product in R3 . In this way dyads and dyadics are nothing but (0, 2)-tensors on R3 ; that is, e1 e2 = e1 ⊗ e2 ∈ T20 (R3 ), by identifying (R3 )∗ with R3 . G. Higher order tensors arise in elasticity and Riemannian geometry. In elasticity, the stress tensor is a symmetric 2-tensor and the elasticity tensor is a fourth-order tensor (see Marsden and Hughes [1983]). In Riemannian geometry the metric tensor is a symmetric 2-tensor and the curvature tensor is a fourth-order tensor. 

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Interior Product. The interior product of a vector v ∈ E (resp., a form β ∈ E∗ ) with a tensor t ∈ Tsr (E; F) is the (r, s−1) (resp., (r−1, s)) type F-valued tensor defined by (iv t)(β 1 , . . . , β r , v1 , . . . , vs−1 ) = t(β 1 , . . . , β r , v, v1 , . . . , vs−1 ) (iβ t)(β 1 , . . . , β r−1 , v1 , . . . , vs ) = t(β, β 1 , . . . , β r−1 , v1 , . . . , vs ). r Clearly, iv : Tsr (E; F) → Ts−1 (E; F) and iβ : Tsr (E; F) → Tsr−1 (E; F) are linear continuous maps, as are β v 7→ iv and β 7→ i . If F = R and dim(E) = n, these operations take the following form in components. If ek (resp., ek ) denotes the kth basis (resp., dual basis) element of E, we have

iek (ei1 ⊗ · · · ⊗ eir ⊗ ej1 ⊗ · · · ⊗ ejs ) = δkj1 ei1 ⊗ · · · ⊗ eir ⊗ ej2 ⊗ · · · ⊗ ejs , k

ie (ei1 ⊗ · · · ⊗ eir ⊗ ej1 ⊗ · · · ⊗ ejs ) = δik1 ei2 ⊗ · · · ⊗ eir ⊗ ej1 ⊗ · · · ⊗ ejs . By Proposition 6.1.2 these formulas and linearity enable us to compute any interior product. Contractions. Let dim(E) = n. The contraction of the kth contravariant with the `th covariant index, r−1 (E) defined for any or for short, the (k, `)-contraction, is the family of linear maps C`k : Tsr (E; F) → Ts−1 pair of natural numbers r, s ≥ 1 by j1 jr s C`k (tij11...i ...jr ei1 ⊗ · · · ⊗ eis ⊗ e ⊗ · · · ⊗ e ) i ...i

pi

...i

s k+1 = tj11 ...jk−1 e ⊗ · · · ⊗ eˆik ⊗ · · · ⊗ eis ⊗ ej1 ⊗ · · · ⊗ eˆj` ⊗ · · · ⊗ ejr , `−1 pj`+1 ...jr i1

where {e1 , . . . , en } is a basis of E, {e1 , . . . , en } is the dual basis in E∗ , and ˆ over a vector or covector means that it is omitted. It is straightforward to verify that C`k so defined is independent of the basis. This is essentially the same computation that is needed to show that the trace of a linear transformation of E to itself is intrinsic. If E is infinite dimensional, contraction is not defined for arbitrary tensors. One introduces the so-called contraction class tensors, analogous to the trace class operators, defines contraction as above in terms of a Banach space basis and its dual, and shows that the contraction class condition implies that the definition is basis independent. We shall not dwell upon these technicalities, refer to Rudin [1973] for a brief discussion of trace class operators, and invite the reader to model the concept of contraction class along these lines. For example, if E∗ = E = `2 (R), and ei = ei equals the sequence with 1 in the ith place and zero everywhere else, then ∞ ∞ X X t= 2−n en ⊗ en ∈ T11 (E) and C11 (t) = 2−n = 2. n=0

n=0

Kronecker Delta. The Kronecker delta is the tensor δ ∈ T11 (E) defined by δ(α, e) = hα, ei. If E is finite dimensional, δ corresponds to the identity I ∈ L(E; E) under the canonical isomorphism T11 (E) ∼ = L(E; E). Relative to any basis, the components of δ are the usual Kronecker symbols δji , that is, δ = δji ei ⊗ ej . Associated Tensors. Suppose E is a finite-dimensional real inner product space with a basis {e1 , . . . , en } and corresponding dual basis {e1 , . . . , en } in E∗ . Using the inner product, with matrix denoted by [gij ], so gij = hhei , ej ii, we get the isomorphism [

The matrix of

[

: E → E∗ given by x 7→ hhx, ·ii ,

and its inverse ] : E∗ → E.

is [gij ]; that is, (x[ )i = gij xj

and of

]

is [g ij ]; that is, (α] )i = g ij αj ,

where xj and αj are the components of x and α, respectively. We call [ the index lowering operator and the index raising operator .

]

6.1 Tensors on Linear Spaces

345

These operators can be applied to tensors to produce new ones. For example if t is a tensor of type (0, 2) we can define an associated tensor t0 of type (1, 1) by t0 (e, α) = t(e, α] ). The components are (t0 )ji = g jk tik tji

(as usual, sum on k).

jk

In the classical literature one writes for g tik , and this is indeed a convenient notation in calculations. However, contrary to the impression one may get from the classical theory of Cartesian tensors, t and t0 are different tensors. 6.1.4 Examples. {e1 , . . . , en }. A.

Let E be a finite-dimensional real vector space with basis {e1 , . . . , en } and dual basis

If t ∈ T12 (E) and x = xi ei , then j p k` j ix t = xp iep (tk` j ek ⊗ e` ⊗ e ) = x tj iep (ek ⊗ e` ⊗ e ) j p k` = xp tk` j δp ek ⊗ e` = x tp ek ⊗ e` .

p Thus, the components of ix t are xp tk` p . The interior product of the same tensor with α = αp e takes the form p` ep j k` p j j iα t = αp tk` j i (ek ⊗ e` ⊗ e ) = αp tj δk e` ⊗ e = αp tj e` ⊗ e .

B.

If t ∈ T32 (E), the (2, 1)-contraction is given by ij k ` m ` m C12 (tij k`m ei ⊗ ej ⊗ e ⊗ e ⊗ e ) = tj`m ei ⊗ e ⊗ e .

C. An important particular example of contraction is the trace of a (1, 1)-tensor. Namely, if t ∈ T11 (E), then trace(t) = C11 (t) = tii , where t = tij ei ⊗ ej . In the remaining examples, assume that there is an inner product on E, denoted, as above, by g(u, v) = hhu, vii. D.

The components of the tensor associated to g by raising the second index are g jk gik = g jk gki = δij .

E.

Let t ∈ T23 (E),

` m t = tijk `m ei ⊗ ej ⊗ ek ⊗ e ⊗ e .

Then t has quite a few associated tensors, depending on which index is lowered or raised. For example tijkm = g mp tijk ` `p , mc iab tim t`c jk` = gja gkb g

tk`m = gia gjb g `c g md tabk ij cd `c ajb tj` ikm = gia gkb g tcm

and so on. F. The positioning of the indices in the components of associated tensors is important. For example, if t ∈ T 02 (E), we saw earlier that tji = g jk tik . However, tji = g ij tki , which is in general different from tji when t is not symmetric. For example, if E = R3 with gij = δij and the nine components of t in the standard basis are t12 = 1, t21 = −1, tij = 0 for all other pairs i, j, then tij = tij , tji = tji , so that t12 = t12 = 1 while t21 = t21 = −1.

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6. Tensors

G. The trace of a (2, 0)-tensor is defined to be the trace of the associated (1, 1) tensor; that is, if t = tij ei ⊗ ej , then trace(t) = tii = gik tik . The question naturally arises whether we get the same answer by lowering the first index instead of the second, that is, if we consider tii . By symmetry of gij we have tii = gki tik = tkk , so that the definition of the trace is independent of which index is lowered. Similarly, if t ∈ T20 (E),

trace(t) = tii = g ik tik = tkk .

In particular, trace(g) = gii = g ik gik = dim(E).



The Dual of a Linear Transformation. If ϕ ∈ L(E, F), the transpose or dual of ϕ, denoted ϕ∗ ∈ L(F∗ , E∗ ) is defined by hϕ∗ (β), ei = hβ, ϕ(e)i, where β ∈ F∗ and e ∈ E. Let us analyze the matrices of ϕ and ϕ∗ in the finite dimensional case. As customary in linear algebra, vectors in a given basis are represented by a column whose entries are the components of the vector. Let ϕ ∈ L(E, F) and let {e1 , . . . , en } and {f1 , . . . , fm } be ordered bases of E and F respectively. Put ϕ(ei ) = Aai fa . (We use a different dummy index for the F-index to avoid confusion.) This defines the matrix of ϕ; A = [Aai ]. Thus, for v = v i ei ∈ E the components of ϕ(v) are given by ϕ(v)a = Aai v i . Hence, thinking of v and ϕ(v) as column vectors, this formula shows that ϕ(v) is computed by multiplying v on the left by A, the matrix of ϕ, as in elementary linear algebra. Thus, the upper index is the row index, while the lower index is the column index. Consequently, ϕ(ei ) represents the ith column of the matrix of ϕ. Let

us now investigate the matrix of ϕ∗ ∈ L(F∗ , E∗ ). In the dual ordered bases, hϕ∗ (f a ), ei i = hf a , ϕ(ei )i = f a , Abi fb = Abi δba = Aai , that is, ϕ∗ (f a ) = Aai ei and thus ϕ∗ (f a ) is the ath row of A. Consequently the matrix of ϕ∗ is the transpose of the matrix of ϕ. If β = βa f a ∈ F∗ then ϕ∗ (β) = βa ϕ∗ (f a ) = βa Aai ei , which says that the ith component of ϕ∗ (β) equals ϕ∗ (β)i = βa Aai . Thinking of elements in the dual as rows whose entries are their components in the dual basis, this shows that ϕ∗ (β) is computed by multiplying β on the right by A, the matrix of ϕ, again in agreement with linear algebra. Push-forward and Pull-back. Now we turn to the effect of linear transformations on tensors. We start with an induced map that acts “forward” like ϕ. 6.1.5 Definition. If ϕ ∈ L(E, F) is an isomorphism, define the push-forward of ϕ, T rs ϕ = ϕ∗ ∈ L(T rs (E), T rs (F)) by ϕ∗ t(β 1 , . . . , β r , f1 , . . . , fs ) = t(ϕ∗ (β 1 ), . . . , ϕ∗ (β r ), ϕ−1 (f1 ), . . . , ϕ−1 (fs )), where t ∈ T rs (E), β 1 , . . . , β r ∈ F∗ , and f1 , . . . , fs ∈ F. We leave the verification that ϕ∗ is continuous to the reader. Note that T 01 ϕ = (ϕ−1 )∗ , which maps “forward” like ϕ. If E and F are finite dimensional, then T 10 (E) = E, T 10 (F) = F and we identify ϕ with T 10 ϕ. The next proposition asserts that the push-forward operation is compatible with compositions and the tensor product. 6.1.6 Proposition.

Let ϕ : E → F and ψ : F → G be isomorphisms. Then

(i) (ψ ◦ ϕ)∗ = ψ∗ ◦ ϕ∗ ; (ii) if i : E → E is the identity, then so is i∗ : T rs (E) → T rs (E); (iii) ϕ∗ : T rs (E) → T rs (F) is an isomorphism, and (ϕ∗ )−1 = (ϕ−1 )∗ ; (iv) If t1 ∈ T r1s1 (E) and t2 ∈ T r2s1 (E), then ϕ∗ (t1 ⊗ t2 ) = ϕ∗ (t1 ) ⊗ ϕ∗ (t2 ).

6.1 Tensors on Linear Spaces

Proof.

347

For (i), ψ∗ (ϕ∗ t)(γ 1 , . . . , γ r , g1 , . . . , gs ) = ϕ∗ t(ψ ∗ (γ 1 ), . . . , ψ ∗ (γ r ), ψ −1 (g1 ), . . . , ψ −1 (gs )) = t(ϕ∗ ψ ∗ (γ 1 ), . . . , ϕ∗ ψ ∗ (γ r ), ϕ−1 ψ −1 (g1 ), . . . , ϕ−1 ψ −1 (gs )) = t((ψ ◦ ϕ)∗ (γ 1 ), . . . , (ψ ◦ ϕ)∗ (γ r ), (ψ ◦ ϕ)−1 (g1 ), . . . , (ψ ◦ ϕ)−1 (gs )) = (ψ ◦ ϕ)∗ t(γ 1 , . . . , γ r , g1 , . . . , gs ),

where γ 1 , . . . , γ r ∈ G∗ , g1 , . . . , gs ∈ G, and t ∈ T rs (E). We have used the fact that the transposes and inverses satisfy (ψ ◦ ϕ)∗ = ϕ∗ ◦ ψ ∗ and (ψ ◦ ϕ)−1 = ϕ−1 ◦ ψ −1 , which the reader can easily check. Part (ii) is an immediate consequence of the definition and the fact that i∗ = i and i−1 = i. Finally, for (iii) we have ϕ∗ ◦ (ϕ−1 )∗ = i∗ , the identity on T rs (F), by (i) and (ii). Similarly, (ϕ−1 )∗ ◦ ϕ∗ = i∗ the identity on T rs (E). Hence (iii) follows. Finally (iv) is a straightforward consequence of the definitions.  Since (ϕ−1 )∗ maps “backward” it is called the pull-back of ϕ and is denoted ϕ∗ . The next proposition gives a connection with component notation. 6.1.7 Proposition. Let ϕ ∈ L(E, F) be an isomorphism of finite dimensional vector spaces. Let [Aai ] denote the matrix of ϕ in the ordered bases {e1 , . . . , en } of E and {f1 , . . . , fn } of F, that is, ϕ(ei ) = Aai fa . Denote by [Bai ] the matrix of ϕ−1 , that is, ϕ−1 (fa ) = Bai ei . Then [Bai ] is the inverse matrix of [Aai ] in the sense that Bai Aaj = δji . Let t ∈ T rs (E)

with components

q ∈ T rs (F)

with components

r tji11...i ...js

relative to

{e1 , . . . , en }

relative to

{f1 , . . . , fn }.

and ...ar qba11...b s ∗

Then the components of ϕ∗ t relative to {f1 , . . . , fn } and of ϕ q relative to {e1 , . . . , en } are given respectively by a ...a

j1 js r (ϕ∗ t)b11...bsr = Aai11 . . . Aairr tji11...i ...js Bb1 . . . Bbs bs i1 ir a1 ...ar b1 r (ϕ∗ q)ij11...i ...js = Ba1 . . . Bar qb1 ...bs Aj1 . . . Ajs .

Proof.

We have ei = ϕ−1 (ϕ(ei )) = ϕ−1 (Aai fa ) = Aai ϕ−1 (fa ) = Aai Baj ej ,

whence Baj Aai = δij for all i, j. Similarly, one shows that Abi Bai = δab , so that [Aai ]−1 = [Baj ]. We have ...ar = (ϕ∗ t)(f a1 , . . . , f ar , fb1 , . . . , fbs ) (ϕ∗ t)ab11...b s

= t(ϕ∗ (f a1 ), . . . , ϕ∗ (f ar ), ϕ−1 (fb1 ), ϕ−1 (fbs )) = t(Aai11 ei1 , . . . , Aairr eir , Bbj11 ej1 , . . . , Bbjss eis ) js r j1 = Aai11 . . . Aairr tji11...i ...js bb1 . . . bbs .

To prove the second relation, we need the matrix of (ϕ−1 )∗ ∈ L(E∗ , F∗ ). We have

−1 ∗ i 



 (ϕ ) (e ), fa = ei , ϕ−1 (fa ) = ei , Bak ek = Bai so that (ϕ−1 )∗ (ei ) = Bai f a . Now proceed as in the previous case.



Note that the matrix of (ϕ−1 )∗ ∈ L(E∗ , F∗ ) is the transpose of the inverse of the matrix of ϕ. The assumption that ϕ be an isomorphism for ϕ∗ to exist is quite restrictive but clearly cannot be weakened. However, one might ask if instead of “push-forward,” the “pull-back” operation is considered, this restrictive assumption can be dropped. This is possible when working with covariant tensors, even when ϕ ∈ L(E, F) is arbitrary.

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6.1.8 Definition. If ϕ ∈ L(E, F) (not necessarily an isomorphism), define the pull-back ϕ∗ ∈ L(T 0s (F), T 0s (E)) by ϕ∗ t(e1 , . . . , es ) = t(ϕ(e1 ), . . . , ϕ(es )), where t ∈ T 0s (F) and e1 , . . . , es ∈ E. Likewise, one can push forward tensors in T0r (E) even if ϕ is not an isomorphism. The next proposition asserts that ϕ∗ is compatible with compositions and the tensor product. Its proof is almost identical to that of proposition 6.1.6 and is left as an exercise for the reader. 6.1.9 Proposition.

Let ϕ ∈ L(E; F) and ψ ∈ L(F; G).

(i) (ψ ◦ ϕ)∗ = ϕ∗ ◦ ψ ∗ . (ii) If i : E → E is the identity, then so is i∗ ∈ L(Ts0 (E), Ts0 (E)). (iii) If ϕ is an isomorphism, then so is ϕ∗ and ϕ∗ = (ϕ−1 )∗ . (iv) If t1 ∈ Ts01 (F) and t2 ∈ Ts02 (F), then ϕ∗ (t1 ⊗ t2 ) = (ϕ∗ t1 ) ⊗ (ϕ∗ t2 ). Finally, the components of ϕ∗ t are given by the following. 6.1.10 Proposition. Let E and F be finite-dimensional vector spaces and ϕ ∈ L(E, F). For ordered bases {e1 , . . . , en } of E and {f1 , . . . , fm } of F, suppose that ϕ(ei ) = Aai fa , and let t ∈ Ts0 (F) have components tb1 ...bs . Then the components of ϕ∗ t relative to {e1 , . . . , en } are given by (ϕ∗ t)j1 ...js = tb1 ...bs Abj11 . . . Abjss . Proof. (ϕ∗ t)j1 ...js = (ϕ∗ t)(ej1 , . . . , ejs ) = t(ϕ(ej1 ), . . . , ϕ(ejs )) = t(Abj11 fb1 , . . . , Abjss fbs ) = t(fb1 , . . . , fbs )Abj11 . . . Abjss = tb1 ...bs Abj11 . . . Abjss



6.1.11 Examples. A. On R2 with the standard basis {e1 , e2 }, let t ∈ T02 (R2 ) be given by t = e1 ⊗e1 +2e1 ⊗e2 −e2 ⊗e1 +3e2 ⊗e2 and let ϕ ∈ L(R2 , R2 ) have the matrix   2 1 A= . 1 1 Then ϕ is clearly an isomorphism, since A has an inverse matrix given by   1 −1 B= . −1 2 According to Proposition 6.1.7, the components of T = ϕ∗ t relative to the standard basis of R2 are given by T ij = Bai Bbj tab ,

with B11 = 1, B12 = B21 = −1, and B22 = 2

6.1 Tensors on Linear Spaces

349

so that T 12 = B11 B12 t11 + B11 B22 t12 + B21 B12 t21 + B21 B22 t22 = 1 · (−1) · 1 + 1 · 2 · 2 + (−1) · (−1) · (−1) + (−1) · 2 · 3 = −4, T 21 = B12 B11 t11 + B12 B21 t12 + B22 B11 t21 + B22 B21 t22 = (−1) · 1 · 1 + (−1) · (−1) · 2 + 2 · 1 · (−1) + 2 · (−1) · 3 = −7, T 11 = B11 B11 t11 + B11 B21 t12 + B21 B11 t21 + B21 B21 t22 = 1 · 1 · 1 + 1 · (−1) · 2 + (−1) · 1 · (−1) + (−1) · (−1) · 3 = 3, T 22 = B12 B12 t11 + B12 B22 t12 + B22 B12 t21 + B22 B22 t22 = (−1) · (−1) · 1 + (−1) · 2 · 2 + 2 · (−1) · (−1) + 2 · 2 · 3 = 11. Thus, ϕ∗ t = 3e1 ⊗ e1 − 4e1 ⊗ e2 − 7e2 ⊗ e1 + 11e2 ⊗ e2 . B. Let t = e1 ⊗ e2 − 2e2 ⊗ e2 ∈ T11 (R2 ) and consider the same map ϕ ∈ L(R2 , R2 ) as in part (a) above. We could compute the components of ϕ∗ t relative to the standard basis of R2 using the formula in Proposition 6.1.7 as before. An alternative way to proceed directly using Proposition 6.1.6(iv), that is, the fact that ϕ∗ is compatible with tensor products. Thus ϕ∗ t = ϕ∗ (e1 ⊗ e2 − 2e2 ⊗ e2 ) = ϕ(e1 ) ⊗ ϕ∗ (e2 ) − 2ϕ(e2 ) ⊗ ϕ∗ (e2 ). But ϕ(e1 ) = 2e1 + e2 , ϕ(e2 ) = e1 + e2 , and ϕ∗ (e2 ) = −e1 + 2e2 , so that ϕ∗ t = (2e1 + e2 ) ⊗ (−e1 + 2e2 ) − 2(e1 + e2 ) ⊗ (−e1 + 2e2 ) = e2 ⊗ e1 − 2e2 ⊗ e2 . C. Let e1 , e2 be the standard basis of R2 and e1 , e2 be the dual basis, as usual. Let t = −2e1 ⊗ e2 ∈ T20 (R2 ) and ϕ ∈ L(R3 , R2 ) be given by the matrix   1 0 2 A= . 0 −1 1 We will compute ϕ∗ t ∈ T20 (R3 ) by using the fact that ϕ∗ is compatible with tensor products and that the matrix of ϕ∗ ∈ L(R2 , R3 ) is the transpose of A . Recall that ϕ∗ (ei ) is the ith row , since matrices act on the right on covectors. Let f 1 , f 2 , f 3 denote the standard dual basis of R3 . Then ϕ∗ (e1 ) = f 1 + 2f 3 and ϕ∗ (e2 ) = −f 2 + f 3 , so that ϕ∗ (t) = −2ϕ∗ (e1 ) ⊗ ϕ∗ (e2 ) = −2(f 1 + 2f 3 ) ⊗ (−f 2 + f 3 ) = 2f 1 ⊗ f 2 − 2f 1 ⊗ f 3 + 4f 3 ⊗ f 2 − 4f 3 ⊗ f 3 .

Exercises  6.1-1.

Compute the interior product of the tensor t = e1 ⊗ e1 ⊗ e2 + 3e2 ⊗ e2 ⊗ e1

with e = −e1 + 2e2 and α = 2e1 + e2 . What are the (1, 1) and (2, 1) contractions of t?



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6. Tensors

 6.1-2. Compute all associated tensors of t = e1 ⊗ e2 ⊗ e2 + 2e2 ⊗ e1 ⊗ e2 − e2 ⊗ e2 ⊗ e1 with respect to the standard metric of R2 .  6.1-3.

Let t = 2e1 ⊗ e1 − e2 ⊗ e1 + 3e1 ⊗ e2 and ϕ ∈ L(R2 , R2 ), ψ ∈ L(R3 , R2 ) be given by the matrices     2 1 0 1 −1 , . −1 1 1 0 2

Compute: trace(t), ϕ∗ t, ψ ∗ t, trace(ϕ∗ t), trace(ψ ∗ t), ϕ∗ t, and all associated tensors of t, ϕ∗ t, ψ ∗ t, and ϕ∗ t with respect to the corresponding standard inner products in R2 and R3 .  6.1-4. Let dim(E) = n and dim(F) = m. Show that Tsr (E; F) is an mnr+s -dimensional real vector space by exhibiting a basis.

6.2

Tensor Bundles and Tensor Fields

We now extend the tensor algebra to local vector bundles, and then to vector bundles. For U ⊂ E (open) recall that U × F is a local vector bundle. Then U × Tsr (F) is also a local vector bundle in view of Proposition 6.1.2. Suppose ϕ : U × F → U 0 × F0 is a local vector bundle mapping and is an isomorphism on each fiber ; that is ϕu = ϕ|{u} × F ∈ L(F, F0 ) is an isomorphism. Also, let ϕ0 denote the restriction of ϕ to the zero section. Then ϕ induces a mapping of the local tensor bundles as follows. 6.2.1 Definition. If ϕ : U × F → U 0 × F0 is a local vector bundle mapping such that for each u ∈ U , ϕu is an isomorphism, let ϕ∗ : U × Tsr (F) → U 0 × Tsr (F0 ) be defined by ϕ∗ (u, t) = (ϕ0 (u), (ϕu )∗ t), where t ∈ Tsr (F). Before proceeding, we shall pause and recall some useful facts concerning linear isomorphisms that were already proved in Lemmas 2.5.4 and 2.5.5. 6.2.2 Proposition. L(E, F) is open.

Let GL(E, F) denote the set of linear isomorphisms from E to F. Then GL(E, F) ⊂

6.2.3 Proposition.

Define the maps A : L(E, F) → L(F∗ , E∗ );

ϕ 7→ ϕ∗

and I : GL(E, F) → GL(F, E);

ϕ 7→ ϕ−1 .

Then A and I are of class C ∞ and DI −1 (ϕ) · ψ = −ϕ−1 ◦ ψ ◦ ϕ−1 . Smoothness of A is clear since it is linear. 6.2.4 Proposition. If ϕ : U × F → U 0 × F0 is a local vector bundle map and ϕu is an isomorphism for all u ∈ U , then ϕ∗ : U × Tsr (F) → U 0 × Tsr (F0 ) is a local vector bundle map and (ϕu )∗ = (ϕ∗ )u is an isomorphism for all u ∈ U . Moreover, if ϕ is a local vector bundle isomorphism then so is ϕ∗ . Proof. That ϕ∗ is an isomorphism on fibers follows from Proposition 6.1.6(iii) and the last assertion follows from the former. By Definition 6.2.1 we need only establish that (ϕu )∗ = (ϕ∗ )u is of class C ∞ . Now ϕu is a smooth function of u, and, by Proposition 6.2.3 ϕ∗u and ϕ−1 u are smooth functions of u. The map (ϕu )∗ is a Cartesian product of r factors ϕ∗u and s factors ϕ−1 , so is smooth. Hence, from the product rule, u (ϕu )∗ is smooth. 

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351

This smoothness can be verified also for finite-dimensional bundles by using the standard bases in the tensor spaces as local bundle charts and proving that the components of ϕ∗ t are C ∞ functions. We have the following commutative diagram, which says that ϕ∗ preserves fibers: U × Tsr (F)

ϕ∗

- U 0 × T r (F0 ) s π0 ? - U0

π ? U ϕ0 Tensor Bundles.

With the above preliminaries out of the way, we can now define tensor bundles.

6.2.5 Definition. b ∈ B. Define

Let π : E → B be a vector bundle with Eb = π −1 (b) denoting the fiber over the point Tsr (E) =

[

Tsr (Eb )

b∈B

and

πsr

:

Tsr (E)

→ B by

πsr (e)

= b where e ∈

Tsr (Eb ).

Furthermore, for a given subset A of B, we define [ Tsr (E)|A = Tsr (Eb ). b∈A

If π 0 : E 0 → B 0 is another vector bundle and (ϕ, ϕ0 ) : E → E 0 is a vector bundle mapping with ϕb = ϕ|Eb an isomorphism for all b ∈ B, let ϕ∗ : Tsr (E) → Tsr (E 0 ) be defined by ϕ∗ |Tsr (Eb ) = (ϕb )∗ . Now suppose that (E|U, ϕ) is an admissible local bundle chart of π, where U ⊂ B is an open set. Then the mapping ϕ∗ |[Tsr (E)|U ] is obviously a bijection onto a local bundle, and thus is a local bundle chart. Further, (ϕ∗ )b = (ϕb )∗ is a linear isomorphism, so this chart preserves the linear structure of each fiber. We shall call such a chart a natural chart of Tsr (E). 6.2.6 Theorem. If π : E → B is a vector bundle, then the set of all natural charts of πsr : Tsr (E) → B is a vector bundle atlas. Proof. Condition VB1 is obvious. For VB2, suppose we have two overlapping natural charts, ϕ∗ and ψ∗ . For simplicity, let them have the same domain. Then α = ψ ◦ ϕ−1 is a local vector bundle isomorphism, and by Proposition 6.1.6, ψ∗ ◦ (ϕ∗ )−1 = α∗ , is a local vector bundle isomorphism by Proposition 6.2.4.  This atlas of natural charts called the natural atlas of πsr , generates a vector bundle structure, and it is easily seen that the resulting vector bundle is Hausdorff, and all fibers are isomorphic Banachable spaces. Hereafter, Tsr (E) will denote all of this structure. 6.2.7 Proposition. If f : E → E 0 is a vector bundle map that is an isomorphism on each fiber, then f∗ : Tsr (E) → Tsr (E 0 ) is also a vector bundle map that is an isomorphism on each fiber. Proof. Let (U, ϕ) be an admissible vector bundle chart of E, and let (V, ψ) be one of E 0 so that f (U ) ⊂ V and fϕψ = ψ ◦ f ◦ ϕ−1 is a local vector bundle mapping. Then using the natural atlas, we see that (f∗ )ϕ∗ ,ψ∗ = (fϕψ )∗ .



6.2.8 Proposition. Let f : E → E 0 and g : E 0 → E 00 be vector bundle maps that are isomorphisms on each fiber. Then so is g ◦ f , and

352

6. Tensors

(i) (g ◦ f )∗ = g∗ ◦ f∗ ; (ii) if i : E → E is the identity, then i∗ : Tsr (E) → Tsr (E) is the identity; (iii) if f : E → E 0 is a vector bundle isomorphism, then so if f∗ and (f∗ )−1 = (f −1 )∗ . Proof. For (i) we examine representatives of (g ◦ f )∗ and g∗ ◦ f∗ . These representatives are the same in view of Proposition 6.1.6. Part (ii) is clear from the definition, and (iii) follows from (i) and (ii) by the same method as in Proposition 6.1.6.  We now specialize to the case where π : E → B is the tangent vector bundle of a manifold. 6.2.9 Definition. Let M be a manifold and τM : T M → M its tangent bundle. We call Tsr (M ) = Tsr (T M ) the vector bundle of tensors contravariant of order r and covariant of order s, or simply of type (r, s). We identify T01 (M ) with T M and call T10 (M ) the cotangent bundle of M also denoted by ∗ τM : T ∗ M → M . The zero section of Tsr (M ) is identified with M . Tensor Fields. Recall that a section of a vector bundle assigns to each base point b a vector in the fiber over b and the addition and scalar multiplication of sections takes place within each fiber. In the case of Tsr (M ) these vectors are called tensors. The C ∞ sections of π : E → B were denoted Γ∞ (π), or Γ∞ (E). Recall that F(M ) denotes the set of mappings from M into R that are of class C ∞ (the standard local manifold structure being used on R) together with its structure as a ring; namely, f + g, cf , f g for f, g ∈ F(M ), c ∈ R are defined by (f + g)(x) = f (x) + g(x),

(cf )(x) = c(f (x)),

and (f g)(x) = f (x)g(x).

Finally, recall that a vector field on M is an element of X(M ) = Γ∞ (T M ). 6.2.10 Definition. A tensor field of type (r, s) on a manifold M is a C ∞ section of Tsr (M ). We denote by Tsr (M ) the set Γ∞ (Tsr (M )) together with its (infinite-dimensional ) real vector space structure. A covector field or differential one-form is an element of X∗ (M ) = T10 (M ). If f ∈ F(M ) and t ∈ Tsr (M ), let f t : M → Tsr (M ) be defined by m 7→ f (m)t(m). If Xi ∈ X(M ), 0 i = 1, . . . , s, αj ∈ X∗ (M ), j = 1, . . . , r, and t0 ∈ Tsr0 (M ) define t(α1 , . . . , αr , X1 , . . . , Xs ) : M → R and

0

r+r t ⊗ t0 : M → Ts+s 0 (M )

6.2.11 Proposition.

m 7→ t(m)(α1 (m), . . . , Xs (m))

by

m 7→ t(m) ⊗ t0 (m).

With f , t, Xi , αj , and t0 as in Definition 6.2.10, f t ∈ Tsr (M ),

Proof.

by

t(α1 , . . . , Xs ) ∈ F(M ),

and

0

r+r t ⊗ t0 ∈ Ts+s 0 .

The differentiability is evident in each case from the product rule in local representation.



For the tangent bundle T M , a natural chart is obtained by taking T ϕ, where ϕ is an admissible chart of M . This in turn induces a chart (T ϕ)∗ on Tsr M . We shall call these the natural charts of Tsr M . Coordinate Representation of Tensor Fields. Recall that ∂/∂xi = (T ϕ)−1 (ei ), for ϕ : U → U 0 ⊂ Rn a chart on M , is a basis of X(U ). The vector field ∂/∂xi corresponds to the derivation f 7→ ∂f /∂xi . Since dxi (∂/∂xj ) = ∂xi /∂xj = δji , we see that dxi is the dual basis of ∂/∂xi at every point of U , that is, that dxi = ϕ∗ (ei ), where {e1 , . . . , en } is the dual basis to {e1 , . . . , en }. Let   ∂ ∂ i1 ir r tij11...i = t dx , . . . , dx , , . . . , ∈ F(U ). ...js ∂xj1 ∂xjs

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353

Applying Proposition 6.1.6(iv) at every point yields the coordinate expression of an (r, s)-tensor field: r t U = tij11...i ...js

∂ ∂ ⊗ · · · ⊗ ir ⊗ dxj1 ⊗ · · · ⊗ dxjs . ∂xi1 ∂x

To discuss the behavior of these components relative to a change of coordinates, assume that X i : U → R, i = 1, . . . , n is a different coordinate system. We can write ∂/∂xi = aji ∂/∂X j , since both are bases of X(U ). Applying both sides to X k yields aji = ∂X j /∂xi that is, (∂/∂xi ) = (∂X j /∂xi )(∂/∂X j ). Thus the dxi , as dual basis change with the inverse of the Jacobian matrix [∂X j /∂xi ]; that is, dxi = (∂xi /∂X j )dX j . Writing t in both coordinate systems and isolating equal terms gives the following change of coordinate formula for the components: ∂X k1 ∂X kr ∂xj1 ∂xjs i1 ...ir ...kr T`k11...` = ··· ··· t i i ` s 1 r 1 ∂x ∂x ∂X ∂X `s j1 ...js r This formula is known as the tensoriality criterion: A set of nr+s functions tji11...i ...js defined for each coordinate system on the open set U of M locally define an (r, s)-tensor field iff changes of coordinates have the aforementioned effect on them. This statement is clear since at every point it assures that the nr+s functions are the components of an (r, s)-tensor in Tu U and conversely. The algebraic operations on tensors, such as contraction, inner products and traces, all carry over fiberwise to tensor fields. For example, if δm ∈ T11 (Tm M ) is the Kronecker delta, then δ : M → T11 (M ); m 7→ δm is obviously C ∞ , and δ ∈ T11 (M ) is called the Kronecker delta. Similarly, a tensor field of type (0, s) or (r, 0) is called symmetric, if it is symmetric at every point.

Metric Tensors.

A basic example of a symmetric covariant tensor field is the following.

6.2.12 Definition. A weak pseudo-Riemannian metric on a manifold M is defined to be a tensor field g ∈ T20 (M ) that is symmetric and weakly nondegenerate, that is, such that at each m ∈ M , g(m)(vm , wm ) = 0 for all wm ∈ Tm M implies vm = 0. A strong pseudo-Riemannian metric is a 2-tensor field that, in addition is strongly nondegenerate for all m ∈ M ; that is, the map vm 7→ g(m)(vm , ·) is an isomorphism ∗ of Tm M onto Tm M . A weak (resp., strong) pseudo-Riemannian metric is called weak (resp., strong ) Riemannian if in addition g(m)(vm , vm ) > 0 for all vm ∈ Tm M , vm 6= 0. It is not hard to show that a strong Riemannian manifold (that is, a manifold with a strong Riemannian metric) is necessarily modeled on a Hilbertizable space; that is, the model space has an equivalent norm arising from an inner product. For finite-dimensional manifolds weak and strong metrics coincide: indeed ∗ ∗ M is an isomorphism. It M have the same dimension and so a one-to-one map of Tm M to Tm Tm M and Tm is possible to have weak metrics on a Banach or Hilbert manifold that are not strong. For example, the L2 inner product on M = C 0 ([0, 1], R) is a weak metric that is not strong. For a similar Hilbert space example, see Exercise 6.2-3. Any Hilbert space is a Riemannian manifold with a constant metric equal to the inner product. A symmetric bilinear (weakly) nondegenerate two-form on any Banach space provides an example of a (weak) pseudo-Riemannian constant metric. A pseudo-Riemannian manifold used in the theory of special relativity is R4 with the Minkowski pseudo-Riemannian metric g(x)(v, w) = v 1 w1 + v 2 w2 + v 3 w3 − v 4 w4 , where x, v, w ∈ R4 . Raising and Lowering Indices. As in the algebraic context of §5.1, pseudo-Riemannian metrics (and for that matter any strongly nondegenerate bilinear tensor) can be used to define associated tensors. Thus the maps ] , [ become vector bundle isomorphisms over the identity [ : T M → T ∗ M , ] : T ∗ M → T M ; ] is [ the inverse of [ , where vm = g(m)(vm , ·). In particular, they induce isomorphisms of the spaces of sections [ ∗ ] : X(M ) → X (M ), : X∗ (M ) → X(M ). In finite dimensions this is the operation of raising and lowering indices. Thus formulas like the ones in Example 6.1.4E should be read pointwise in this context.

354

6. Tensors

Gradients.

There is a particular index raising operation that requires special attention.

6.2.13 Definition. Let M be a pseudo-Riemannian n-manifold with metric g. For f ∈ F(M ), the vector field defined by grad f = (df )] ∈ X(M ) is called the gradient of f . To find the expression of grad f in local coordinates, we write   ∂ ∂ ∂ ∂ , , X = X i i , and Y = Y i i , gij = g ∂xi ∂xj ∂x ∂x so we have D

[

X ,Y

E

i

j



= g(X, Y ) = X Y g

∂ ∂ , j i ∂x ∂x



= X i Y j gij ; that is, X [ = X i gij dxj . If α ∈ X∗ (M ) has the coordinate expression α = αi dxi , we have α] = αi g ij ∂/∂xj where [g ij ] is the inverse of the matrix [gij ]. Thus for α = df , the local expression of the gradient is grad f = g ij

∂f ∂ ; ∂xj ∂xi

that is,

i

(grad f ) = g ij

∂f . ∂xj

If M = Rn with standard Euclidean metric gij = δij , this formula becomes grad f =

∂f ∂ ; ∂xi ∂xi

 that is,

grad f =

∂f ∂f ,..., n ∂x1 ∂x

 ,

the familiar expression of the gradient from vector calculus. Push-forward and Pull-back of Tensor Fields. Now we turn to the effect of mappings and diffeomorphisms on tensor fields. As motivation, we mention a way that the pull-back comes up in applications. Consider the deformation of an elastic body, thought of as a mapping (usually an embedding) ϕ : B → R3 , where B ⊂ R3 (say an open set) represents a reference configuration, such as an unstressed state of the body. Letting g be the usual Riemannian structure on R3 , then ϕ∗ g, which measures the way that ϕ distorts lengths and angles, is called the Cauchy-Green tensor and it measures the strain in the material. See Marsden and Hughes [1983] for further elaboration. 6.2.14 Definition. If ϕ : M → N is a diffeomorphism and t ∈ Tsr (M ), let ϕ∗ t = (T ϕ)∗ ◦ t ◦ ϕ−1 , be the push-forward of t by ϕ. If t ∈ Tsr (N ), the pull-back of t by ϕ is given by ϕ∗ t = (ϕ−1 )∗ t. 6.2.15 Proposition.

If ϕ : M → N is a diffeomorphism, and t ∈ Tsr (M ), then

(i) ϕ∗ t ∈ Tsr (N ); (ii) ϕ∗ : Tsr (M ) → Tsr (N ) is a linear isomorphism; (iii) (ϕ ◦ ψ)∗ = ϕ∗ ◦ ψ∗ ; and (iv) ϕ∗ (t ⊗ t0 ) = ϕ∗ t ⊗ ϕ∗ t0 , where t ∈ Tsr (M ) and t0 ∈ Tsr (M ). Proof. (i) The differentiability is evident from the composite mapping theorem, together with Proposition 6.2.4. The other three statements are proved fiberwise, where they are consequences of Proposition 6.1.6.  As in the algebraic context, the pull-back of covariant tensors is defined even for maps that are not diffeomorphisms. Globalizing Definition 6.1.8 we get the following.

6.2 Tensor Bundles and Tensor Fields

6.2.16 Definition.

355

If ϕ : M → N and t ∈ Ts0 (N ), then ϕ∗ t, the pull-back of t by ϕ, is defined by (ϕ∗ t)(m)(v1 , . . . , vs ) = t(ϕ(m))(Tm ϕ(v1 ), . . . , Tm ϕ(vs ))

for m ∈ M, v1 , . . . , vs ∈ Tm M . The next proposition is similar to Proposition 6.2.15 and is proved by globalizing the proof of Proposition 6.1.9. 6.2.17 Proposition.

If ϕ : M → N is C ∞ and t ∈ Ts0 (N ), then

(i) ϕ∗ ∈ Ts0 (M ); (ii) ϕ∗ : Ts0 (N ) → Ts0 (M ) is a linear map; (iii) (ψ ◦ ϕ)∗ = ϕ∗ ◦ ψ ∗ for ψ : N → P ;
∗ (iv) if ϕ is a diffeomorphism them ϕ∗ is an isomorphism with inverse ϕ−1 ; and (v) t1 ∈ Ts01 (N ), t2 ∈ Ts02 (N ), then ϕ∗ (t1 ⊗ t2 ) = (ϕ∗ t1 ) ⊗ (ϕ∗ t2 ). For finite-dimensional manifolds the coordinate expressions of the pull-back and push-forward can be read directly from Propositions 6.1.7 and 6.1.10, taking into account that T ϕ is given locally by the Jacobian matrix. This yields the following. 6.2.18 Proposition. Let M and N be finite-dimensional manifolds, ϕ : M → N a C r map and denote by y j = ϕj (x1 , . . . , xm ) the local expression of ϕ relative to charts where m = dim(M ) and j = 1, . . . , n = dim(N ). (i) If t ∈ Tsr (M ) and ϕ is a diffeomorphism, the coordinates of the push-forward ϕ∗ t are  i1   ir  `1 ∂y ∂y ∂x ∂x`s k1 ...kr i ...i −1 −1 (ϕ∗ t)j11 ...jrs = ◦ ϕ · · · ◦ ϕ . . . t ◦ ϕ−1 . k k j ∂x 1 ∂x r ∂y 1 ∂y js `1 ...`s If t ∈ Ts0 (N ) and ϕ is a diffeomorphism, the coordinates of the pull-back ϕ∗ t are  i1   ir  k1 ∂x ∂x ∂y ∂y ks r r = ◦ ϕ . . . ◦ ϕ · · · js tk`11...` (ϕ∗ t)ij11...i ...js ...ks ◦ ϕ. ` ` j 1 r 1 ∂y ∂y ∂x ∂x (ii) If t ∈ Ts0 (N ) and ϕ : M → N is arbitrary, the coordinates of the pull-back ϕ∗ t are (ϕ∗ t)j1 ...js =

∂y k1 ∂y ks · · · tk ...k ◦ ϕ. ∂xj1 ∂xjs 1 s

Notice the similarity between the formulas for coordinate change and pull-back. The situation is similar to the passive and active interpretation of similarity transformations PAP−1 in linear algebra. Of course it is important not to confuse the two. 6.2.19 Examples. A.

Let ϕ : R2 → R2 be defined by ϕ(x, y) = (x + 2y, y) and let    
∂ ∂ t = 3x ⊗ dy + ⊗ dy ∈ T11 R2 . ∂x ∂y

The matrix of ϕ∗ on vector fields is 

  ∂ϕi 1 = 0 ∂xj



2 1

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6. Tensors

and on forms is 

∂xi ∂ϕj

 =

 1 0

−1

2 1

=

  1 −2 . 0 1

In other words,  ϕ∗

∂ ∂x





∂ = , ∂x

ϕ∗

ϕ∗ (dx) = dx − 2 dy,

∂ ∂y

 =2

∂ ∂ + . ∂x ∂y

ϕ∗ (dy) = dy.

Noting that ϕ−1 (x, y) = (x − 2y, y), we get  ϕ∗ t = 3(x − 2y)ϕ∗

∂ ∂x





∂ ∂y



⊗ ϕ∗ (dy) + ϕ∗ ⊗ ϕ∗ (dy)   ∂ ∂ ∂ = 3(x − 2y) ⊗ dy + 2 + ⊗ dy ∂x ∂x ∂y ∂ ∂ ⊗ dy + ⊗ dy. = (3x − 6y + 2) ∂x ∂y

B.

With the same mapping and tensor, we compute ϕ∗ t. Since     ∂ ∂ ∂ ∂ ∂ ∗ ∗ = , ϕ = −2 + . ϕ ∂x ∂x ∂y ∂x ∂y ϕ∗ (dx) = dx + 2 dy,

ϕ∗ (dy) = dy,

we have ∗



∂ ∂x



∗

∗



∂ ∂y



⊗ ϕ (dy) + ϕ ⊗ ϕ∗ (dy)   ∂ ∂ ∂ ⊗ dy + −2 + ⊗ dy = 3(x + 3y) ∂x ∂x ∂y ∂ ∂ = (3x + 6y − 2) ⊗ dy + ⊗ dy. ∂x ∂y

ϕ t = 3(x + 2y)ϕ

C.

∗

Let ϕ : R3 → R2 , ϕ(x, y, z) = (2x + z, xyz) and t = (u + 2v)du ⊗ du + (u)2 du ⊗ dv ∈ T20 (R2 ). Since ϕ∗ (du) = 2dx + dz

and ϕ∗ (dv) = yzdx + xzdy + xydz,

we have ϕ∗ t = (2x + z + 2xyz)(2dx + dz) ⊗ (2 dx + dz) + (2x + z)2 (2 dx + dz) ⊗ (yz dx + xz dy + xy dz) = 2[4x + 2z + 4xyz + (2x + z)2 yz] dx ⊗ dx + 2(2x + z)2 xz dx ⊗ dy + 2[2x + z + 2xyz + (2x + z)2 xy] dx ⊗ dz + [4x + 2z + 4xyz + yz(2x + z)2 ]dz ⊗ dx + xz(2x + z)2 dz ⊗ dy + [2x + z + 2xyz + xy(2x + z)2 ]dz ⊗ dz.

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357

D. If ϕ : M → N represents the deformation of an elastic body and g is a Riemannian metric on N , then C = ϕ∗ g is called the Cauchy–Green tensor ; in coordinates Cij =

∂ϕα ∂ϕβ gαβ ◦ ϕ. ∂xi ∂xj

Thus, C measures how ϕ deforms lengths and angles.



Alternative Approach to Tensor Fields. Suppose that F(M ) and X(M ) have been defined. With the “scalar multiplication” (f, X) 7→ f X defined in Definition 6.2.10, X(M ) becomes an F(M )-module. That is, X(M ) is essentially a vector space over F(M ), but the “scalars” F(M ) form only a commutative ring with identity, rather than a field. Define LF (M ) (X(M ), F(M )) = X ∗ (M ) the F(M )-linear mappings on X(M ), and similarly ∗ Trs (M ) = Lr+s F (M ) (X (M ), . . . , X(M ); F(M ))

the F(M )-multilinear mappings. From Definition 6.2.10, we have a natural mapping Tsr (M ) → Trs (M ) which is F(M )-linear. 6.2.20 Proposition. Let M be a finite-dimensional manifold or be modeled on a Banach space with norm C ∞ away from the origin. Then Tsr (M ) is isomorphic to Trs (M ) regarded as F(M )-modules and as real vector spaces. In particular, X∗ (M ) is isomorphic to X ∗ (M ). Proof.

Consider the map Tsr (M ) → Trs (M ) given by `(α1 , . . . , αr , X1 , . . . , Xs )(m) = `(m)(α1 (m), . . . , Xs (m)).

This map is clearly F(M )-linear. To show it is an isomorphism, given such a multilinear map `, define t by

t (m) α1 (m) , . . . , Xs (m) = ` α1 , . . . , Xs (m) . To show that t is well-defined we first show that, for each v0 ∈ Tm M , there is an X ∈ X(M ) such that X(m) = v0 , and similarly for dual vectors. Let (U, ϕ) be a chart at m and let Tm ϕ(v0 ) = (ϕ(m), v00 ). Define Y ∈ X(U 0 ) by Y (u) = (u0 , v00 ) on a neighborhood V1 of ϕ(m), where w = ϕ(n). Extend Y to U 0 so Y is zero outside V2 , where cl(V1 ) ⊂ V2 , cl(V2 ) ⊂ U 0 , by means of a bump function. Define X by Xϕ = Y on U , and X = 0 outside U . Then X(m) = v0 . The construction is similar for dual vectors. As in Theorem 4.2.16, F(M )-linearity of ` shows that the definition of t(m) is independent of how the vectors v0 (and corresponding dual vectors) are extended to fields. The tensor field t(m) so defined is C ∞ ; indeed, using the chart ϕ, the local representative of t is C ∞ by Supplement 3.4A, since ` induces a C ∞ map M × Tsr (M ) → R (by the composite function theorem), which is (r + s)-linear at every m ∈ M . If M is finite dimensional this last step of the proof can be simplified as follows. In the chart ϕ with coordinates (x1 , . . . , xn ), ∂ ∂ r t = tij11...i ⊗ · · · ⊗ ir ⊗ dxj1 ⊗ · · · ⊗ dxjs ...js ∂xi1 ∂ and all components of t are C ∞ by hypothesis.  The preceding proposition can be clearly generalized to the C k situation. One can also get around the use of a smooth norm on the model space if one assumes that the multilinear maps are localizable, that is, are defined on X∗ (U ) × · · · × X(U ) with values in F(U ) for any open set U in a way compatible with restriction to U . We shall take this point of view in the next section.

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6. Tensors

The direct sum T (M ) of the Tsr (M ), including T00 (M ) = F(M ), is a real vector space with ⊗-product, called the tensor algebra of M , and if ϕ : M → N is a diffeomorphism, ϕ∗ : T (M ) → T (N ) is an algebra isomorphism. The construction of T (M ) and the properties discussed in this section can be generalized to vector bundle valued (r, s)-tensors (resp. tensor fields), that is, elements (resp. sections) of L(T ∗ M ⊕ · · · ⊕ T ∗ M ⊕ T M ⊕ · · · ⊕ T M, E), the vector bundle of vector bundle maps from T ∗ M ⊕ · · · ⊕ T M (with r factors of T ∗ M and s factors of T M ) to the vector bundle E, which cover the identity map of the base M .

Exercises  6.2-1.

Let ϕ : R2 \{ (0, y) | y ∈ R } → R2 \{ (x, x) | x ∈ R } be defined by ϕ(x, y) = (x3 + y, y) and let t=x

∂ ∂ ⊗ dx ⊗ dy + y ⊗ dy ⊗ dy. ∂x ∂y

Show that ϕ is a diffeomorphism and compute ϕ∗ t, ϕ∗ t. Endow R2 with the standard Riemannian metric. Compute the associated tensors of t, ϕ∗ t, and ϕ∗ t as well as their (1, 1) and (1, 2) contractions. What is the trace of the interior product of t with ∂/∂x + x∂/∂y?  6.2-2. Let ϕ : R2 → R3 , ϕ(x, y) = (y, x, y + x2 ) be the deformation of an elastic shell. Compute the Cauchy–Green tensor and its trace.  6.2-3.

Let H be the set of real sequences {an }n=1,2,... such that X kan k2 = n2 a2n < ∞. n≥1

Show that H is a Hilbert space. Show that g (a, b) =

X

an bn

n≥1

is a weak Riemannian metric on H that is not a strong metric.  6.2-4.

Let (M, g) be a Riemannian manifold and let N ⊂ M be a submanifold. Define νg (N ) = { v ∈ Tn M | g(n)(v, u) = 0 for allu ∈ Tn N and all n ∈ N }.

Show that νg (N ) is a sub-bundle of T M |N isomorphic to both the normal and conormal bundles ν(N ) and µ(N ) defined in Exercises 3.4-10 and 3.4-11.  6.2-5. (i) Let E be a vector space and hh , ii an inner product. If the associated flat map ϕ : E → E∗ is onto, then show that E is a Hilbert space. (ii) Formulate a theorem for a weak Riemannian manifold to be a strong Riemannian manifold using the result in (i).

6.3

The Lie Derivative: Algebraic Approach

This section extends the Lie derivative £X from vector fields and functions to the full tensor algebra. We shall do so in two ways. This section does this algebraically and in the next section, it is done in terms of the flow of X. The two approaches will be shown to be equivalent.

6.3 The Lie Derivative: Algebraic Approach

359

Differential Operators. We shall demand certain properties of £X such as: if t is a tensor field of type (r, s), so is £X t, and £X should be a derivation for tensor products and contractions. First of all, how should £X be defined on covector fields? If Y is a vector field and α is a covector field, then the contraction α · Y is a function, so £X (α · Y ) and £X Y are already defined. (See §4.2.) However, if we require the derivation property for contractions, namely £X (α · Y ) = (£X α) · Y + α · (£X Y ), then this forces us to define £X α by (£X α) · Y = £X (α · Y ) − α · (£X Y ) for all vector fields Y . Since this defines an F(M )-linear map, £X α is a well-defined covector field. The extension to general tensors now proceeds inductively in the same spirit. 6.3.1 Definition. A differential operator on the full tensor algebra T (M ) of a manifold M is a collection {Dsr (U )} of maps of Tsr (U ) into itself for each r and s ≥ 0 and each open set U ⊂ M , any of which we denote merely D(the r, s and U are to be inferred from the context), such that D01. D is a tensor derivation, or D commutes with contractions, that is, D is R-linear and if t ∈ Tsr (M ), α1 , . . . , αr ∈ X∗ (M ), and X1 , . . . , Xs ∈ X(M ), then D(t(α1 , . . . , αr , X1 , . . . , Xs )) = (Dt) (α1 , . . . , αr , X1 , . . . , Xs ) r X + t(α1 , . . . , Dαj , . . . , αr , X1 , . . . , Xs ) +

j=1 s X

t (α1 , . . . , αr , X1 , . . . , DXk , . . . , Xs ) .

k=1

D02. D is local , or is natural with respect to restrictions. That is, for U ⊂ V ⊂ M open sets, and t ∈ Tsr (V ) (Dt)|U = D(t|U ) ∈ Tsr (U ) that is, the following diagram commutes Tsr (V )

|U

D ? Tsr (V )

|U

- T r (U ) s D ? - Tsr (U )

We do not demand that D be natural with respect to push-forward by diffeomorphisms. Indeed, several important differential operators, such as the covariant derivative, are not natural with respect to diffeomorphisms, although the Lie derivative is, as we shall see. 6.3.2 Theorem.

Suppose for each open set U ⊂ M we have maps EU : F(U ) → F(U )

and

FU : X(U ) → X(U ),

which are (R-linear) tensor derivations and natural with respect to restrictions. That is

360

6. Tensors

(i) EU (f ⊗ g) = (EU f ) ⊗ g + f ⊗ EU g for f, g ∈ F(U ); (ii) for f ∈ F(M ), EU (f |U ) = (EM f )|U ; (iii) FU (f ⊗ X) = (EU f ) ⊗ X + f ⊗ FU X for f ∈ F(U ), and X ∈ X(U ); (iv) for X ∈ X(M ), FU f (X|U ) = (FM X)|U . Then there is a unique differential operator D on T (M ) that coincides with EU on F(U ) and with FU on X(U ). Proof.

Since D must be a tensor derivation, define D on X∗ (U ) by the formula (Dα) · X = D(α · X) − α · (DX) = EU (α · X) − α · FU X

for all X ∈ X(U ). By properties (i) and (iii), Dα is F(M )-linear and thus by the remark following Proposition 6.2.20, D so defined on X∗ (U ) has values in X∗ (U ). Note also that D(f ⊗ α) = (Ef ) ⊗ α + f ⊗ (Dα) for any α ∈ X∗ (U ), f ∈ F(U ). This shows that D exists and is unique on X∗ (U ) (by the Hahn–Banach theorem). Define DU on Tsr (U ) by requiring D01 to hold: (DU t)(α1 , . . . , αr , X1 , . . . , Xs ) = EU (t(α1 , . . . , αr , X1 , . . . , Xs )) r X t(α1 , . . . , Dαj , . . . , αr , X1 , . . . , Xs ) − j=1

−

s X

t(α1 , . . . αr , X1 , . . . , FU Xk , . . . , Xs ).

k=1

From (i), (iii), and D01 for DU on X∗ (U ), it follows that DU t is an F(M )-multilinear map, that is that DU t ∈ Tsr (U ) (see the comment following Proposition 6.2.20). The definition of DU on Tsr (U ) uniquely determines DU from the property D01. Finally, if V is any open subset of U , by (ii) and (iv) it follows that DV (t|V ) = (DU t) |V. This enables us to define D on F(M ) by (Dt)(m) = (DU t)(m), where U is any open subset of M containing m. Since DU is unique, so is D, and so D02 is satisfied by the construction of D.  6.3.3 Corollary.

We have

(i) D(t1 ⊗ t2 ) = Dt1 ⊗ t2 + t1 ⊗ Dt2 , and (ii) Dδ = 0, where δ is Kronecker’s delta. Proof. (i) is a direct application of D01. For (ii) let α ∈ X∗ (U ) and X ∈ X(U ) where U is an arbitrary chart domain. Then (Dδ)(α, X) = D(δ(α, X)) − δ(Dα, X) − δ(α, DX) = D(α · X) − Dα · X − α · DX = 0. Again the Hahn–Banach theorem assures that Dδ = 0 on U , and thus by D02, Dδ = 0.



6.3 The Lie Derivative: Algebraic Approach

361

The Lie Derivative. Taking EU and FU to be £X|U we see that the hypotheses of Theorem 6.3.2 are satisfied. Hence we can define a differential operator as follows. 6.3.4 Definition. If X ∈ X(M ), we let £X be the unique differential operator on T (M ), called the Lie derivative with respect to X, such that £X coincides with £X as given on F(M ) and X(M ) (see Definitions 4.2.6 and 4.2.20). 6.3.5 Proposition. Let ϕ : M → N be a diffeomorphism and X a vector field on M . Then £X is natural with respect to push-forward by ϕ; that is, £ϕ∗ X ϕ∗ t = ϕ∗ £X t

for Tsr (M ),

or the following diagram commutes: ϕ∗

Tsr (M ) £X ? Tsr (M )

Proof.

ϕ∗

- T r (N ) s £ϕ∗ X ? - Tsr (N )

For an open set U ⊂ M define D : Tsr (U ) → Tsr (U )

by

Dt = ϕ∗ £ϕ∗ X|U (ϕ∗ t),

where we use the same symbol ϕ for ϕ|U . By naturality on F(U ) and X(U ), D coincides with £X|U on F(U ) and X(U ). Next, we show that D is a differential operator. For D01, we use the fact that ϕ∗ (t(α1 , . . . , αr , X1 , . . . , Xs )) = (ϕ∗ t)(ϕ∗ α1 , . . . , ϕ∗ αr , ϕ∗ X1 , . . . , ϕ∗ Xs ), which follows from the definitions. Then for X, X1 , . . . , Xs ∈ X(U ) and α1 , . . . , αr ∈ X∗ (U ), D(t(α1 , . . . , αr , X1 , . . . , Xs )) = ϕ∗ £ϕ∗ X (ϕ∗ (t(α1 , . . . , αr , X1 , . . . , Xs ))) = ϕ∗ £ϕ∗ X ((ϕ∗ t)(ϕ∗ α1 , . . . , ϕ∗ αr , ϕ∗ X1 , . . . , ϕ∗ Xs )) = ϕ∗ [(£ϕ∗ X ϕ∗ t)(ϕ∗ α1 , . . . , ϕ∗ αr , ϕ∗ X1 , . . . , ϕ∗ Xs )) r X + (ϕ∗ t)(ϕ∗ α1 , . . . , £ϕ∗ X ϕ∗ αj , . . . , ϕ∗ αr , ϕ∗ X1 , . . . , ϕ∗ Xs ) j=1

+

s X

(ϕ∗ t)(ϕ∗ α1 , . . . , ϕ∗ αr , ϕ∗ X1 , . . . , £ϕ∗ X ϕ∗ Xk , . . . , ϕ∗ Xs )],

k=1

by D01 for £X . Since ϕ∗ = (ϕ−1 )∗ by Definition 6.2.14, the preceding expression becomes (Dt)(α1 , . . . αr , X1 , . . . , Xs ) + +

r X j=1 s X

t(α1 , . . . , Dαj , . . . , αr , X1 , . . . , Xs ) t(α1 , . . . , αr , X1 , . . . , DXk , . . . , Xs ).

k=1

For D02, let t ∈ Tsr (M ) and write Dt|U = [(ϕ∗ )−1 £ϕ∗ X ϕ∗ t]|U = (ϕ∗ )−1 [£ϕ∗ X ϕ∗ t]|U = (ϕ∗ )−1 £ϕ∗ X|U ϕ∗ t|U (by D02 for £X ) = D(t|U ).

362

6. Tensors

The result now follows by Theorem 6.3.2.



Using the same reasoning, a differential operator that is natural with respect to diffeomorphisms on functions and vector fields is natural on all tensors. Local Formula for the Lie Derivative. Let us now compute the local formula for £X t where t is a tensor field of type (r, s). Let ϕ : U ⊂ M → V ⊂ E be a local chart and let X 0 and t0 be the principal parts of the local representatives, ϕ∗ X and ϕ∗ t respectively. Thus X 0 : V → E and t0 : V → Tsr (E). Recall from equations (4.2.2) and (4.2.9) that the local formulas for the Lie derivatives of functions and vector fields are: (£X f )0 (x) = Df 0 (x) · X 0 (x)

(6.3.1)

where f 0 is the local representative of f and (£X Y )0 (x) = DY 0 (x) · X 0 (x) − DX 0 (x) · Y 0 (x).

(6.3.2)

Recall also from equations (4.2.3) and (4.2.10) that in finite dimensions, these formulas become £X f = X i

∂f ∂xi

(6.3.3)

and [X, Y ]i = X j

i ∂Y i j ∂X − Y . ∂xj ∂xj

(6.3.4)

Let us first find the local expression for £X α where α is a one-form. By Proposition 6.3.5, the local representative of £X α is ϕ∗ (£X α) = £ϕ∗ X ϕ∗ α, which we write as £X 0 α0 where X 0 and α0 are the principal parts of the local representatives, so X 0 : V → E and α0 : V → E∗ . Let v ∈ E be fixed and regarded as a constant vector field. Then as £X is a tensor derivation, £X 0 (α0 · v) = (£X 0 α0 ) · v + α0 · (£X 0 v). By equations (6.3.1) and (6.3.2) this becomes D(α0 · v) · X 0 = (£X 0 α0 ) · v − α0 · (DX 0 · v). Thus, (£X 0 α0 ) · v = (Dα0 · X 0 ) · v + α0 · (DX 0 · v). In the expression (Dα0 · X 0 ) · v, Dα0 · X 0 means the derivative of α0 in the direction X 0 ; the resulting element of E∗ is then applied to v. Thus we can write £X 0 α0 = Dα0 · X 0 + α0 · DX 0 .

(6.3.5)

In finite dimensions, the corresponding coordinate expression is (£X α)i v i =

∂αi j i ∂X j i X v + αj v; j ∂x ∂xi

that is, (£X α)i = X j

∂αi ∂X j + α . j ∂xj ∂xi

(6.3.6)

6.3 The Lie Derivative: Algebraic Approach

363

Now let t be of type (r, s), so t0 : V → L(E∗ , . . . , E∗ , E, . . . , E; R). Let α1 , . . . , αr be (constant) elements of E∗ and v1 , . . . , vs (constant) elements of E. Then again by the derivation property, £X 0 [t0 (α1 , . . . , αr , v1 , . . . , vs )] = (£X 0 t0 )(α1 , . . . , αr , v1 , . . . , vs ) r X + t0 (α1 , . . . , £X 0 αi , . . . , αr , v1 , . . . , vs ) i=1

+

s X

t0 (α1 , . . . , αr , v1 , . . . , £X 0 vj , . . . , vs ).

j=1

Using the local formulas (6.3.1)–(6.3.5) for the Lie derivatives of functions, vector fields, and one-forms, we get (Dt0 · X 0 ) · (α1 , . . . , αr , v1 , . . . , vs ) = (£X 0 t0 )(α1 , . . . , αr , v1 , . . . , vs ) r X + t0 (α1 , . . . , αi · DX 0 , . . . , αr , v1 , . . . , vs ) i=1

+

s X

t0 (α1 , . . . , αr , v1 , . . . , −DX 0 · vj , . . . , vs ).

j=1

Therefore, (£X 0 t0 )(α1 , . . . , αr , v1 , . . . , vs ) = (Dt0 · X 0 )(α1 , . . . , αr , v1 , . . . , vs ) r X − t0 (α1 , . . . , αi · DX 0 , . . . , αr , v1 , . . . , vs ) i=1

+

s X

t0 (α1 , . . . , αr , v1 , . . . , DX 0 · vj , . . . , vs ).

j=1

In components, this reads ∂X i1 `i2 ...ir ∂ i1 ...ir tjs ...js − t − (all upper indices) k ∂x ∂x` j1 ...js ∂X m i1 ...ir + t + (all lower indices) ∂xj1 mj2 ...js

k r (£X t)ij11...i ...js = X

(6.3.7)

We deduced the component formulas for £X t in the case of a finite-dimensional manifold as corollaries of the general Banach manifold formulas. Because of their importance, we shall deduce them again in a different manner, without appealing to Proposition 6.3.5. Let r t = tij11...i ...js

∂ ∂ ⊗ · · · ⊗ ir ⊗ dxj1 · · · ⊗ dxjs ∈ Tsr (U ), ∂xi1 ∂x

where U is a chart domain on M . If X = X k ∂/∂xk , the tensor derivation property can be used to compute £X t. For this we recall that on scalars, k r £X (tji11...i ...js ) = X

r ∂tji11...i ...js ∂xk

364

6. Tensors

and that

  ∂ ∂X i ∂ ∂ £X k = X, k = − k ∂x ∂x ∂x ∂xi

by the general formula for the bracket components. The formula for £X (dxk ) is found in the following way. The relation δik = dxk (∂/∂xi ) implies by D01 that        ∂ ∂ ∂ k k 0 = £X dxk = (£ (dx + dx )) X, X ∂xi ∂xi ∂xi     ∂ ∂X ` ∂ = (£X (dxk )) + dxk − i . i ∂x ∂x ∂x` Thus, (£X (dxk ))



∂ ∂xi



= dxk

so £X (dxk ) =





∂X ` ∂ ∂xi ∂x`

∂X k ∂xi



 =

∂X k , ∂xi

dxi .

Now one simply applies D01 and collects terms to get the same local formula for £X t found in equation (6.3.7). Note especially that   ∂ ∂ ∂ £ i = 0 and £ i (dxj ) = 0, for all i, j. ∂x ∂xj ∂x 6.3.6 Examples. A.

Compute £X t, where t=x

∂ ∂ ⊗ dx ⊗ dy + y ⊗ dy ⊗ dy ∂y ∂y

and

X=

∂ ∂ +x . ∂x ∂y

Solution Method 1. First of all, note that £X t = £(∂/∂x+x∂/∂y) t = £∂/∂x t + £x∂/∂y t The first term on the right hand side is   ∂ ∂ £∂/∂x t = £∂/∂x x ⊗ dx ⊗ dy + y ⊗ dy ⊗ dy ∂y ∂y     ∂ ∂ = £∂/∂x x ⊗ dx ⊗ dy + £∂/∂x y ⊗ dy ⊗ dy ∂y ∂y ∂ = ⊗ dx ⊗ dy + 0. ∂y Similarly, the second term on the right hand side of (6.3.8) is   ∂ ∂ £x∂/∂y t = £x∂/∂y x ⊗ dx ⊗ dy + y ⊗ dy ⊗ dy ∂y ∂y   ∂ = 0+0+0+x ⊗ dx ⊗ dx ∂y   ∂ ∂ ∂ + x ⊗ dy ⊗ dy + 0 + y ⊗ dx ⊗ dy + y ⊗ dy ⊗ dx . ∂y ∂y ∂y

(6.3.8)

(6.3.9)

(6.3.10)

6.3 The Lie Derivative: Algebraic Approach

365

Thus, substituting (6.3.9) and (6.3.10) into (6.3.8), we find ∂ ∂ ∂ ⊗ dx ⊗ dy + x ⊗ dx ⊗ dx + x ⊗ dy ⊗ dy ∂y ∂y ∂y ∂ ∂ +y ⊗ dx ⊗ dy + y ⊗ dy ⊗ dx ∂y ∂y ∂ ∂ ∂ = (y + 1) ⊗ dx ⊗ dy + x ⊗ dx ⊗ dx + x ⊗ dy ⊗ dy ∂y ∂y ∂y ∂ +y ⊗ dy ⊗ dx. ∂y

£X t =

Method 2. Using component notation, t is a tensor of type (1, 2) whose nonzero components are t212 = x and t222 = y. The components of X are X 1 = 1 and X 2 = x. Thus, by the component formula (6.3.7), (£X t)ijk = X m

∂X i ∂X m ∂X p ∂ i tjk − t`jk ` + timk + tijp k . m j ∂x ∂x ∂x ∂x

The nonzero components are (£X t)212 = 1 − 0 + y + 0 = 1 + y;

(£X t)222 = x − 0 + 0 + 0 = x;

(£X t)211 = 0 − 0 + 0 + x = x;

(£X t)221 = 0 − 0 + 0 + y = y,

and hence ∂ ∂ ⊗ dx ⊗ dy + x ⊗ dx ⊗ dx ∂y ∂y ∂ ∂ +x ⊗ dy ⊗ dy + y ⊗ dy ⊗ dx. ∂y ∂y

£X t = (y + 1)

The two methods thus give the same answer. It is useful to understand both methods since they both occur in the literature, and depending on the circumstances, one may be easier to apply than the other. B. In Riemannian geometry, vector fields X satisfying £X g = 0 are called Killing vector fields; their geometric significance will become clear in the next section. For now, let us compute the system of equations that the components of a Killing vector field must satisfy. If X = X i ∂/∂xi , and g = gij dxi ⊗ dxj , then £X g = (£X gij )dxi ⊗ dxj + gij (£X dxi ) ⊗ dxj + gij dxi ⊗ (£X dxj ) ∂X i ∂X j k ∂gij = X k k dxi ⊗ dxj + gij k dxk ⊗ dxj + gij dxi ⊗ dx ∂x ∂x ∂xk   ∂gij ∂X k ∂X k = X k k + gkj + g dxi ⊗ dxj . ik ∂x ∂xi ∂xj Note that £X g is still a symmetric (0, 2)-tensor, as it must be. Hence X is a Killing vector field iff its components satisfy the following system of n partial differential equations, called Killing’s equations Xk

∂X k ∂X k ∂gij + gkj + gik = 0. k i ∂x ∂x ∂xj

C. In the theory of elasticity, if u represents the displacement vector field , the expression £u g is called the strain tensor . As we shall see in the next section, this is related to the Cauchy–Green tensor C = ϕ∗ g by linearization of the deformation ϕ.

366

6. Tensors

D. Let us show that £X does not necessarily commute with the formation of associated tensors; for example, that (£X t)ij 6= (£X τ )ij , where t = tij ∂/∂xi ⊗ dxj ∈ T11 (M ) and τ = tij dxi ⊗ dxj ∈ T20 (M ) is the associated tensor with components tij = gik tkj . We have from equation (6.3.7) (£X t)ij = X k

i k ∂tij k ∂X i ∂X − t + t , j k ∂k k ∂xk ∂xj

and so (£X t)ij = gi`

` k ∂t`j k ∂X ` ∂X − t + t X j k ∂xk ∂xk ∂xj k

! .

From equation (6.3.7), ∂X ` ∂X k ∂tij + t` j + tik k i ∂x ∂x ∂xj k ∂ ∂X ` ` ∂X = X k k (gi` t`j ) + g`k tkj + g t i` k ∂x ∂xi ∂xj ` ` ∂tj ∂gi` ∂X ∂X k = X k k t`j + X k gi` k + g`k tkj + gi` t`k . i ∂x ∂x ∂x ∂xj

(£X τ )ij = X k

Thus, to have equality it is necessary and sufficient that Xk

` ` ∂gi` ` k ∂X k ∂X t + g t + g t =0 `k i` j j ∂xk j ∂xi ∂xk

for all pairs of indices (i, j), which is a nontrivial system of n2 linear partial differential equations for gij . If X is a Killing vector field, then ∂X ` ∂X ` ∂gik g`k gi` k = −X ` ` , ∂xi ∂x ∂x which substituted in the preceding equation, gives zero. The converse statement is proved along the same lines. In other words, a necessary and sufficient condition that £X commute with the formation of associated tensors is that X be a Killing vector field for the pseudo-Riemannian metric g.  As usual, the development of £X extends from tensor fields to F -valued tensor fields.

Exercises  6.3-1.

Let t = xy


∂ ∂ ∂ ⊗ dx + y ⊗ dx + ⊗ dy ∈ T11 R2 . ∂x ∂y ∂x

Define the map ϕ as follows: ϕ : { (x, y) | y > 0 } → { (x, y) | x > 0, x2 < y } and ϕ(x, y) = (yex , y 2 e2x + y). Show that ϕ is a diffeomorphism and compute trace(t), ϕ∗ t, ϕ∗ t, £X t, £X ϕ∗ t, and Lϕ∗ X t, for X = y∂/∂x + x2 ∂/∂y.  6.3-2. Verify explicitly that £X (t[ ) 6= (£X t)[ where lowered, for X and t in Exercise 6.3-1.

[

denotes the associated tensor with both indices

 6.3-3. Compute the coordinate expressions for the Killing equations in R3 in rectangular, cylindrical, and spherical coordinates. What are the Killing vector fields in Rn ?  6.3-4. Let (M, g) be a finite dimensional pseudo-Riemannian manifold, and g ] the tensor g with both indices raised. Let X ∈ X(M ). Calculate (£X g ] )[ − £X g in coordinates.

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 6.3-5. If (M, g) is a finite-dimensional pseudo-Riemannian manifold and f ∈ F(M ), X ∈ X(M ), calculate £X (∇f ) − ∇(£X f ).  6.3-6. (i) Let t ∈ T11 (M ). Show that there is a unique tensor field Nt ∈ T21 (M ), skew–symmetric in its covariant indices, such that £t·X t − t · £X t = Nt · X for all X ∈ X(M ), where the dots mean contractions, that is (t · X)i = tij X j , (t · s)ij = tik skj , where i i t, s ∈ T11 (M ), and Nt · X = Njk X k , where Nt = Njk ∂/∂xi ⊗ dxj ⊗ dxk . Nt is called the Nijenhuis tensor . Generalize to the infinite-dimensional case. i Hint: Show that Njk = t`k tij,` − t`j tik,` + ti` t`k,j − ti` t`j,k .

(ii) Show that Nt = 0 iff [t · X, t · Y ] − t · [t · X, Y ] = t · [X, t · Y ] − t2 · [X, Y ] for all X, Y ∈ X(M ), where t2 ∈ T11 (M ) is the tensor field obtained by the composition t ◦ t, when t is thought of as a map t : X(M ) → X(M ).

6.4

The Lie Derivative: Dynamic Approach

We now turn to the dynamic interpretation of the Lie derivative. In §4.2 it was shown that £X acting on an element of F(M ) or X(M ), respectively, is the time derivative at zero of that element of F(M ) or X(M ) Lie dragged along by the flow of X. The same situation holds for general tensor fields. Given t ∈ Tsr (M ) and X ∈ X(M ), we get a curve through t(m) in the fiber over m by using the flow of X. The derivative of this curve is the Lie derivative. 6.4.1 Theorem (Lie Derivative Theorem). Consider the vector field X ∈ Xk (M ), the tensor field t ∈ Tsr (M ) both of class C k , k ≥ 1 and Fλ be the flow of X. Then on the domain of the flow (see Figure 5.4.1), we have d ∗ F t = Fλ∗ £X t. dλ λ

m

Figure 6.4.1. The Lie derivative

Proof.

It suffices to show that

d F ∗ t = £X t. dλ λ=0 λ

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Indeed, if this is proved then d d d ∗ ∗ F t = F ∗ F ∗ t = Fλ∗ £X t. Fλ t = dλ dµ µ=0 µ+λ dµ µ=0 λ µ Define θX : T (M ) → T (M ) by d θX (t)(m) = (F ∗ t)(m). dλ λ=0 λ Note that θX (t) is a smooth tensor field of the same type as t, by smoothness of t and Fλ . (We suppress the notational clutter of restricting to the domain of the flow.) Let us apply Theorem 6.3.2. Clearly θX is R-linear and is natural with respect to restrictions. It is a tensor derivation from the product rule for derivatives and the relation (ϕ∗ t)(ϕ∗ α1 , . . . , ϕ∗ αr , ϕ∗ X1 , . . . , ϕ∗ Xs ) = ϕ∗ (t(α1 , . . . , αr , X1 , . . . , Xs )) for ϕ a diffeomorphism. Hence θX is a differential operator. It remains to show that θX coincides with £X on F(M ) and X(M ). For f ∈ F(M ), and X ∈ X(M ), we have d d ∗ F f = £X f and θX Y = F ∗ Y = [X, Y ] θX f = dλ λ=0 λ dλ λ=0 λ by Theorems 4.2.10 and 4.2.19, respectively. By Theorem 6.3.2 and Definition 6.3.4, θX t = £X t for all t ∈ T (M ).  This theorem can also be verified in finite dimensions by a straightforward coordinate computation. See Exercise 6.4-1. The identity in this theorem relating flows and Lie derivatives is so basic, some authors like to take it as the definition of the Lie derivative (see Exercise 6.4-3). 6.4.2 Corollary.

If t ∈ T (M ), £X t = 0 iff t is constant along the flow of X. That is, t = Fλ∗ t.

As an application of Theorem 6.4.1, let us generalize the naturality of £X with respect to diffeomorphisms. As remarked in §5.2, the pull-back of covariant tensor fields makes sense even when the mapping is not a diffeomorphism. It is thus natural to ask whether there is some analogue of Proposition 6.3.5 for pull-backs with no invertibility assumption on the mapping ϕ. Of course, the best one can hope for, since vector fields can be operated upon only by diffeomorphisms, is to replace the pair X, ϕ∗ X be a pair of ϕ-related vector fields. 6.4.3 Proposition. ϕ∗ (£Y t) = £X ϕ∗ t.

Let ϕ : M → N be C ∞ , X ∈ X(M ), Y ∈ X(N ), X ∼ϕ Y , and t ∈ Ts0 (N ). Then

Proof. Recall from Proposition 4.2.4 that X ∼ϕ Y iff Gλ ◦ ϕ = ϕ ◦ Fλ , where Fλ and Gλ are the flows of X and Y , respectively. Thus by Theorem 6.4.1, d d ∗ ∗ ∗ £X (ϕ t) = F ϕ t= (ϕ ◦ Fλ )∗ t dλ λ=0 λ dλ λ=0 d d ∗ (Gλ ◦ ϕ) t = ϕ∗ G∗λ t = dλ λ=0 dλ λ=0 ∗ d =ϕ G∗ t = ϕ∗ (£Y t).  dλ λ=0 λ As with functions and vector fields, the Lie derivative can be generalized to include time-dependent vector fields.

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6.4.4 Theorem (First Time-dependent Lie Derivative Theorem). Let Xλ ∈ Xk (M ), k ≥ 1, for λ ∈ R and suppose that X(λ, m) is continuous in (λ, m). Letting Fλ,µ be the evolution operator for Xλ , we have d ∗ ∗ F t = Fλ,µ (£Xλ t) dλ λ,µ where t ∈ Tsr (M ) is of class C k . Warning. It is not generally true for time-dependent vector fields that the right hand-side in the preceding display equals ∗ £Xλ Fλ,µ t. Proof.

As in Theorem 6.4.1, it is enough to prove the formula at λ = µ where Fλ,λ = identity, for then d d d ∗ ∗ ∗ (F ◦ F ) F∗ t Fλ,µ t = t = F ρ,λ λ,µ λ,µ dλ dρ ρ=λ dρ ρ=λ ρ,λ ∗ = Fλ,µ £Xλ t.

As in Theorem 6.4.1, d θXµ t = F∗ t dλ λ=µ λ,µ is a differential operator that coincides with £Xµ on F(M ) and on X(M ) by Theorem 4.2.31. Thus by Theorem 6.3.2, θXµ = £Xµ on all tensors.  Let us generalize the relationship between Lie derivatives and flows one more step. Call a smooth map t : R × M → Tsr (M ) satisfying tλ (m) = t(λ, m) ∈ (Tm M )rs a time-dependent tensor field . Theorem 6.4.4 generalizes to this context as follows. 6.4.5 Theorem (Second Time-dependent Lie Derivative Theorem). Let tλ be a C k time-dependent tensor, and Xλ be as in Theorem 6.4.4, k ≥ 1, and denote by Fλ,µ the evolution operator of Xλ . Then d ∗ ∗ F tλ = Fλ,µ dλ λ,µ Proof.



∂tλ + £Xλ tλ ∂λ



By the product rule for derivatives and Theorem 6.4.4 we get d d ∂tλ ∗ ∗ ∗ F t = F t + F λ σ σ,µ dλ λ=σ λ,µ dλ λ=σ λ,µ ∂λ λ=σ ∂tλ ∗ ∗ . = Fσ,µ (£Xσ tσ ) + Fσ,µ ∂λ λ=σ



6.4.6 Examples. A. If g is a pseudo-Riemannian metric on M , the Killing equations are £X g = 0 (see Example 6.3.6B). By Corollary 6.4.2 this says that Fλ∗ g = g, where Fλ is the flow of X, that is that the flow of X consists of isometries. B. In elasticity, the vanishing of the strain tensor means, by Example A, that the body moves as a rigid body. 

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The Lie Transform Method. This is an important technique that makes use of the dynamic approach to the Lie derivative, called the Lie transform method . It has been used already in the proof of the Frobenius theorem (§4.4) and we shall see it again in Chapters 7 and 9. The method is also used in the theory of normal forms (cf. Takens [1974], Guckenheimer and Holmes [1983], and Golubitsky and Schaeffer [1985]). 6.4.7 Example (The Lie Transform Method). Let two tensor fields t0 and t1 be given on a smooth manifold M . We say they are locally equivalent at m0 ∈ M if there is a diffeomorphism ϕ of one neighborhood of m0 to another neighborhood of m0 , such that ϕ∗ t1 = t0 . One way to show that t0 and t1 are equivalent is to join them with a curve t(λ) satisfying t(0) = t0 , t(1) = t1 and to seek a curve of local diffeomorphisms ϕλ such that ϕ0 = identity and ϕ∗λ t(λ) = t0 , λ ∈ [0, 1]. If this is done, ϕ = ϕ1 is the desired diffeomorphism. A way to find the curve of diffeomorphisms ϕλ satisfying the relation above is to solve the equation £Xλ t(λ) +

d t(λ) = 0 dλ

for Xλ . If this is possible, let ϕλ = Fλ,0 , where Fλ,µ is the evolution operator of the time-dependent vector field Xλ . Then by Theorem 6.4.5 we have   d ∗ d ϕλ t(λ) = ϕ∗λ £Xλ t(λ) + t(λ) = 0 dλ dλ so that ϕ∗λ t(λ) = ϕ∗0 (0) = t0 . If we choose Xλ so Xλ (m0 ) = 0, then ϕλ exists for a time ≥ 1 by Corollary 4.1.25 and ϕλ (m0 ) = m0 . One often takes t(λ) = (1 − λ)t0 + λt1 . Also, in applications this method is not always used in exactly this way since the algebraic equation for Xλ might be hard to solve. We shall see this happen in the proof of the Poincar´e lemma 7.4.14. The reader should now also look back at the Frobenius theorem 4.4.7 and recognize the spirit of the Lie transform method in its proof.  The Morse Lemma. We now prove a version of the classical Morse lemma in infinite dimensions using the method of Lie transforms. The proof below is due to Golubitsky and Marsden [1983]; see Palais [1969] and Tromba [1976] for the original proofs; Palais’ proof is similar in spirit to the one we give. 6.4.8 Theorem (The Morse–Palais–Tromba Lemma). Let E be a Banach space and h , i a weakly nondegenerate, continuous, symmetric bilinear form on E. Let h : U → R be C k , k ≥ 3, where U is open in E, and let u0 ∈ U satisfy h(u0 ) = 0, Dh(u0 ) = 0. Let B = D2 h(u0 ) : E × E → R. Assume that there is a linear isomorphism T : E → E such that B(u, v) = hT u, vi for all u, v ∈ E and that h has a C k−1 gradient h∇h(y), ui = Dh(y) · u. Then there is a local C k−2 diffeomorphism ϕ of E with ϕ(u0 ) = u0 , Dϕ(u0 ) = I, and h(ϕ(x)) = Proof.

1 B(x − u0 , x − u0 ). 2

Symmetry of B implies that T is self-adjoint relative to h , i. Let   1 f (y) = B (y − u0 , y − u0 ) , 2

h1 = h, and hλ = f + λp, where p(y) = h(y) −

  1 B (y − u0 , y − u0 ) 2

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371

is C k and satisfies p(u0 ) = 0, Dp(u0 ) = 0, and D2 p(u0 ) = 0. We apply the Lie transform method to hλ . Thus we have to solve the following equation for a C k−2 vector field Xλ £Xλ hλ +

dhλ = 0, dλ

Xλ (u0 ) = 0.

(6.4.1)

Then ϕ∗1 h = f , where ϕλ = Fλ,0 , for Fλ,µ the evolution operator of Xλ , and hence ϕ = ϕ1 is a C k−2 diffeomorphism of a neighborhood of u0 satisfying ϕ1 (u0 ) = u0 . If we can prove that Dϕ1 (u0 ) = I, ϕ1 = ϕ will be the desired diffeomorphism. To solve equation (6.4.1), differentiate Dp(x) · e = h∇p(x), ei with respect to x and use the symmetry of the second derivative to conclude that D∇p(x) is symmetric relative to h , i. Therefore,  Z 1 D∇p(u0 + τ (x − u0 )) · (x − u0 )dτ, e Dp(x) · e = h∇p(x), ei = 0

= hT (x − u0 ), R(x) · ei

(6.4.2)

where R : U → L(E, E) is the C k−2 map given by Z 1 R(x) = T −1 Dp(u0 + τ (x − u0 )) · (x − u0 ) dτ 0

which satisfies R(u0 ) = 0. Thus p(y) has the expression Z 1 p(y) = Dp(u0 + τ (y − u0 )) · (y − u0 )dτ = − hT (y − u0 ), X(y)i 0

where X : U → E is the C

k−2

vector field given by Z 1 X(y) = − τ R(u0 + τ (y − u0 )) · (y − u0 ) dτ 0

which satisfies X(u0 ) = 0 and DX(u0 ) = I. Therefore (£Xλ hλ )(y) = Dhλ (y) · Xλ (y) = B(y − u0 , Xλ (y)) + λDp(y) · Xλ (y) = hT (y − u0 ), (I + λR(y)) · Xλ (y)i by (equation 6.4.2) so that the equation (6.4.1) becomes, hT (y − u0 ), (I + λR(y)) · Xλ (y)i = hT (y − u0 ), X(y)i . Since R(u0 ) = 0, there exists a neighborhood of u0 , such that the norm of λR(y) is < 1 for all λ ∈ [0, 1]. Thus for y in this neighborhood, I + λR(y) can be inverted and we can take Xλ (y) = (I + λR(y))−1 X(y) which is a C k−2 vector field defined for all λ ∈ [0, 1] and which satisfies Xλ (u0 ) = 0, DXλ (u0 ) = 0. Differentiating the relation (d/dλ)ϕλ (u) = Xλ (ϕλ (u)) in u at u0 and using ϕλ (u0 ) = u0 yields d Dϕλ (u0 ) = DXλ (ϕλ (u0 )) ◦ Dϕλ (u0 ) = DXλ (u0 ) ◦ Dϕλ (u0 ) = 0, dλ that is Dϕλ (u0 ) is constant in λ ∈ [0, 1]. Since it equals I at λ = 0, it follows that Dϕλ (u0 ) = I.



6.4.9 Corollary (The Classical Morse Lemma). Let h : U → R be C k , k ≥ 3, U open in Rn , and let u ∈ U be a nondegenerate critical point of h, that is h(u) = 0, Dh(u) = 0 and the symmetric bilinear form D2 h(u) on Rn is nondegenerate with index i. Then there is a local C k−2 diffeomorphism ψ of Rn fixing u such that
2
2
2 1h 1 h (ψ(x)) = x − u1 + · · · + xn−i − un−i − xn−i+1 − un−i+1 2 i 2

− · · · − (xn − un )

.

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Proof. In Theorem 6.4.8, take h , i to be the dot-product in Rn to find a local C k−2 diffeomorphism on Rn fixing u0 such that h(ϕ(x)) = (1/2)D2 h(u)(x − u, x − u). Next, apply the Gram–Schmidt procedure to find a basis of Rn in which the matrix of D2 h(u) is diagonal with entries ±1 (see Proposition 7.2.9 for a review of the proof of the existence of such a basis). If i is the number of −1’s (the index ), let ϕ be the composition of ψ with the linear isomorphism determined by the change of an arbitrary basis of Rn to the one above. 

Exercises  6.4-1. Verify Theorem 6.4.1 by a coordinate computation as follows. Let Fλ (x) = (y 1 (λ, x), . . . , y n (λ, x)) so that ∂y i /∂λ = X i (y) and ∂y i /∂xj satisfy the variational equation ∂ ∂y i ∂X i ∂y k = . ∂λ ∂xj ∂xk ∂xj Then write ...ar (Fλ∗ t)ab11...b = s

∂xar ∂y j1 ∂y js i1 ...ir ∂xa1 . · · · · · · t ∂y i1 ∂y ir ∂xb1 ∂xbs j1 ...js

Differentiate this in λ at λ = 0 and obtain the coordinate expression (6.3.7) of §6.3 for £X t.  6.4-2.

Carry out the proof outlined in Exercise 6.4-1 for time-dependent vector fields.

 6.4-3. Starting with Theorem 6.4.1 as the definition of £X t, check that £X satisfies DO1, DO2 and the properties (i)–(iv) of Theorem 6.3.2. r−1 (M ). Use both Theorem 6.4.1 and DO1  6.4-4. Let C be a contraction operator mapping Tsr (M ) to Ts−1 to show that £X (Ct) = C(£X t).

 6.4-5.

Extend Theorem 6.4.1 to F-valued tensors.

 6.4-6. Let f (y) = (1/2)y 2 − y 3 + y 5 . Use the Lie transform method to show that there is a local diffeomorphism ϕ, defined in a neighborhood of 0 ∈ R such that (f ◦ ϕ)(x) = x2 /2.  6.4-7.

Let E = `2 (R), let hx, yi =

∞ X 1 xn yn n n=1

and

h(x) =

∞ ∞ 1X1 2 1X 3 xn = x . 2 n=1 n 3 n=1 n

Show that h vanishes on (0, 0, . . . , 3/2n, 0, . . . ) which → 0 as n → ∞, so the conclusion of the Morse lemma fails. What hypothesis in Theorem 6.4.8 does not hold?  6.4-8 (Buchner, Marsden, and Schecter [1983b]). In the notation of Exercise 2.4-15, show that f has a sequence of critical points approaching 0, so the Morse lemma fails. (The only missing hypothesis is that ∇h is C 1 .)

6.5

Partitions of Unity

A partition of unity is a technical device that is often used to piece smooth local tensor fields together to form a smooth global tensor field. Partitions of unity will be useful for studying integration; in this section they are used to study when a manifold admits a Riemannian metric.

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6.5.1 Definition. If t is a tensor field on a manifold M , the carrier of t is the set of m ∈ M for which t(m) 6= 0, and is denoted carr(t). The support of t, denoted supp(t), is the closure of carr(t). We say t has compact support if supp(t) is compact in M . An open set U ⊂ M is called a C r carrier if there exists an f ∈ F r (M ), such that f ≥ 0 and U = carr(f ). A collection of subsets {Cα } of a manifold M (or, more generally, a topological space) is called locally finite if for each m ∈ M , there is a neighborhood U of m such that U ∩ Cα = ∅ except for finitely many indices α. 6.5.2 Definition.

A partition of unity on a manifold M is a collection {(Ui , gi )}, where

(i) {Ui } is a locally finite open covering of M ; (ii) gi ∈ F(M ), gi (m) ≥ 0 for all m ∈ M , and supp(gi ) ⊂ Ui for all i; (iii) for each m ∈ M , Σi gi (m) = 1. (By (i), this is a finite sum.) If A = {(Vα , ϕα )} is an atlas on M , a partition of unity subordinate to A is a partition of unity {(Ui , gi )} such that each open set Ui is a subset of a chart domain Vα(i) . If any atlas A has a subordinate partition of unity, we say M admits partitions of unity . Occasionally one works with C k partitions of unity. They are defined in the same way except gi are only required to be C k rather than C ∞ . 6.5.3 Theorem (Patching Construction). Let M be a manifold with an atlas A = {(Vα , ϕα )} where ϕα : Vα → Vα0 ⊂ E is a chart. Let tα be a C k tensor field, k ≥ 1, of fixed type (r, s) defined on Vα0 for each α, and assume that there exists a partition of unity {(Ui , gi )} subordinate to A. Let t be defined by t(m) =

X

gi ϕ∗α(i) tα(i) (m),

i

a finite sum at each m ∈ M . Then t is a C k tensor field of type (r, s) on M . Proof. Since {Ui } is locally finite, the sum at every point is a finite sum, and thus t(m) is a type (r, s) tensor for every m ∈ M . Also, t is C k since the local representative of t in the chart (Vα(i) , ϕα(i) ) is Σj (gi ◦ ϕ−1 α(j) )tα(j) , the summation taken over all indices j such that Vα(i) ∩ Vα(j) 6= ∅; by local finiteness the number of these j is finite.  Clearly this construction is not unique; it depends on the choices of the indices α(i) such that Ui ⊂ Vα(i) and on the functions gi . As we shall see later, under suitable hypotheses, one can always construct partitions of unity; again the construction is not unique. The same construction (and proof) can be used to patch together local sections of a vector bundle into a global section when the base is a manifold admitting partitions of unity subordinate to any open covering. To discuss the existence of partitions of unity and consequences thereof, we need some topological preliminaries. 6.5.4 Definition. Let S be a topological space. A covering {Uα } of S is called a refinement of a covering {Vi } if for every Uα there is a Vi such that Uα ⊂ Vi . A topological space is called paracompact if every open covering of S has a locally finite refinement of open sets, and S is Hausdorff. 6.5.5 Proposition.

Second-countable, locally compact Hausdorff spaces are paracompact.

Proof. By second countability S and local compactness of S, there exists a sequence O1 , . . . , On , . . . of open sets with cl(On ) compact and n∈N On = S. Let Vn = O1 ∪ · · · ∪ On , n = 1, 2, . . . and put U1 = V1 . Since {Vn } is an open covering of S and cl(U1 ) is compact, cl(U1 ) ⊂ Vi1 ∪ · · · ∪ Vir .

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Put U2 = Vi1 ∪ · · · ∪ Vir ; then cl(U2 ) is compact. Proceed inductively to show that S is the countable union of open sets Un such that cl(Un ) is compact and cl(Un ) ⊂ Un+1 . If Wα is a covering of S by open sets, and Kn = cl(Un )\Un−1 , then we can cover Kn by a finite number of open sets, each of which is contained in some Wα ∩ Un+1 , and is disjoint from cl(Un−2 ). The union of such collections yields the desired refinement of {Wα }.  Another class of paracompact spaces are the metrizable spaces (see Lemma 6.5.15 in Supplement 6.5A). In particular, Banach spaces are paracompact. 6.5.6 Proposition.

Every paracompact space is normal.

Proof. We first show that if A is closed and u ∈ S\A, there are disjoint neighborhoods of u and A (regularity). For each v ∈ A, let Uu , Vv be disjoint neighborhoods of u and v. Let Wα be a locally finite S refinement of the covering { Vv , S\A | v ∈ A }, and V = Wα , the union over those α with Wα ∩ A 6= ∅. A neighborhood U0 of u meets a finite number of Wα . Let U denote the intersection of U0 and the corresponding Uu . Then V and U are the required neighborhoods. The case for two closed sets proceeds somewhat similarly, so we leave the details for the reader.  Later we shall give general theorems on the existence of partitions of unity. However, there is a simple case that is commonly used, so we present it first. 6.5.7 Theorem. unity.

Let M be a second-countable (Hausdorff ) n-manifold. Then M admits partitions of

Proof. The proof of Proposition 6.5.5 shows the following. Let M be an n-manifold and {Wα } be an open covering. Then there is a locally finite refinement consisting of charts (Vi , ϕi ) such that ϕi (Vi ) is the disk of radius 3, and such that ϕ−1 i (D1 (0)) cover M , where D1 (0) is the unit disk, centered at the origin in the model space. Now let A be an atlas on M and let {(Vi , ϕi )} be a locally finite refinement with these properties. From Lemma 4.2.13, there is a nonzero function hi ∈ F(M ) whose support lies in Vi and hj ≥ 0. Let hi (u) gi (u) = Σi hi (u) (the sum is finite). These are the required functions.



If {Vα } is an open covering of M , we can always find an atlas A = {(Ui , ϕi )} such that {Ui } is a refinement of {Vα } since the atlases generate the topology. Thus, if M admits partitions of unity, we can find partitions of unity subordinate to any open covering. The case of C 0 -partitions of unity differs drastically from the smooth case. Since we are primarily interested in this latter case, we summarize the topological situation, without giving the proofs. 1. If S is a Hausdorff space, the following are equivalent: (i) S is normal . (ii) (Urysohn’s lemma.) For any two closed nonempty disjoint sets A, B there is a continuous function f : S → [0, 1] such that f (A) = 0 and f (B) = 1. (iii) (Tietze extension theorem.) For any closed set A ⊂ S and continuous function g : A → [a, b], there is a continuous extension G : S → [a, b] of g. 2. A Hausdorff space is paracompact iff it admits a C 0 partition of unity subordinate to any open covering.

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It is clear that if {(Ui , gi )} is a continuous partition of unity subordinate to the given open covering {Vα }, then by definition {Ui } is an open locally finite refinement. The converse—the existence of partitions of unity—is the hard part; the proof of this and of the equivalences of (i), (ii), and (iii) can be found for instance in Kelley [1975] and Choquet [1969, Section 6]. These results are important for the rich supply of continuous functions they provide. We shall not use these topological theorems in the rest of the book, but we do want their smooth versions on manifolds. Note that if M is a manifold admitting partitions of unity subordinate to any open covering, then M is paracompact, and thus normal by Proposition 6.5.6. This already enables us to generalize (ii) and (iii) to the smooth (or C k ) situation. 6.5.8 Proposition. Let M be a manifold admitting smooth (or C k ) partitions of unity. If A and B are closed disjoint sets then, there exists a smooth (or C k ) function f : M → [0, 1] such that f (A) = 0 and f (B) = 1. Proof. As we saw, the condition on M implies that M is normal and thus there is an atlas {(Uα , ϕα )} such that Uα ∩ A 6= ∅ implies Uα ∩ B = ∅. Let {(Vi , gi )} be a subordinate C k partition on unity and f = Σgi , where the sum is over those i for which Vi ∩ B 6= ∅. Then f is C k , is one on B, and zero on A.  6.5.9 Theorem (Smooth Tietze Extension Theorem). Let M be a manifold admitting partitions of unity, and let π : E → M be a vector bundle with base M . Suppose σ : A → E is a C k section defined on the closed set A (i.e., every point a ∈ A has a neighborhood Ua and a C k section σa : Ua → E extending σ). Then σ can be extended to a C k global section Σ : M → E. In particular, if g : A → F is a C k function defined on the closed set A, where F is a Banach space, then there is a C k extension G : M → F; if g is bounded by a constant R, that is, kg(a)k ≤ R for all a ∈ A, then so is G. Proof. Consider the open covering { Uα , M \A | a ∈ A } of M , with Uα given by the definition of smoothness on the closed set A. Let {(Ui , gi )} be a C k partition of unity subordinate to this open covering and define σi : Ui → E, by σi = σa |Ui for all Ui and σi ≡ 0 on all Ui disjoint from Ua , a ∈ A. Then gi σi : Ui → E is a C k section on Ui and since supp(gi σi ) ⊂ supp(gi ) ⊂ Ui , it can be extended in a C k manner to M by putting it equal to zero on M \Ui . Thus gi σi : M → E is a C k -section of π : E → M and hence Σ = Σi gi σi is a C k section; note that the sum is finite in a neighborhood of every point m ∈ M . Finally, if a ∈ A ! X X Σ(a) = gi (a)σi (a) = gi (a) σ(a) = σ(a), i

i

that is, Σ|A = σ. The second part of the theorem is a particular case of the one just proved by considering the trivial bundle M × F → M and the section σ defined by σ(m) = (m, g(m)). The boundedness statement follows from the given construction, since all the gi have values in [0, 1].  Construction of Metrics. Before discussing general questions on the existence of partitions of unity on Banach manifolds, we discuss the existence of Riemannian metrics. Recall that a Riemannian metric on a Hausdorff manifold M is a tensor field g ∈ T20 (M ) such that for all m ∈ M , g(m) is symmetric and positive definite. Our goal is to find topological conditions on an n-manifold that are necessary and sufficient to ensure the existence of Riemannian metrics. The proof of the necessary conditions will be simplified by first showing that any Riemannian manifold is a metric space. For this, define for m, n ∈ M , d(m, n) = inf{ `(γ) | γ : [0, 1] → M is a continuous piecewise C 1 curve with γ(0) = m, γ(1) = n }. Here `(γ) is the length of the curve γ, defined by Z

1

kγ(t)k ˙ dt,

`(γ) = 0

(6.5.1)

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where γ(t) ˙ = dγ/dt is the tangent vector at γ(t) to the curve γ and 1/2 kγk ˙ = [gγ(t) (γ(t), ˙ γ(t))] ˙

is its length. 6.5.10 Proposition. The function d defined by equation (6.5.1) is a metric on each connected component of M whose metric topology is the original topology of M . If d is a complete metric, M is called a complete Riemannian manifold . Proof.

Clearly d(m, m) = 0, d(m, n) = d(n, m), and d(m, p) ≤ d(m, n) + d(n, p),

by using the definition. Next we will verify that d(m, n) > 0 whenever m 6= n. Let m ∈ U ⊂ M where (U, ϕ) is a chart for M and suppose ϕ(U ) = U 0 ⊂ E. Then for any u ∈ U , g(u)(v, v)1/2 , defined for v ∈ Tu M , is a norm on Tu M . This is equivalent to the norm on E, under the linear isomorphism Tm ϕ. Thus, if g 0 is the local expression for g, then g 0 (u0 ) defines an inner product on E, yielding equivalent norms for all u0 ∈ U 0 . Using continuity of g and choosing U 0 to be an open disk in E, we can conclude that the norms g 0 (u0 )1/2 and g 0 (m)1/2 , where m0 = ϕ(m) satisfy: 1/2

ag 0 (m0 )

1/2

≤ g 0 (u0 )

1/2

≤ bg 0 (m0 )

for all u0 ∈ U 0 , where a and b are positive constants. Thus, if η : [0, 1] → U 0 is a continuous piecewise C 1 curve, then Z 1 1/2 `(η) = g 0 (η(t))(η(t), ˙ η(t)) ˙ dt 0

Z ≥a

1 1/2 g 0 (m0 )(η(t), ˙ η(t)) ˙ dt

0 0

0

Z

≥ ag (m )

1

Z η(t)dt, ˙

0

1/2

1

η(t)dt ˙ 0

≥ ag 0 (m0 )(η(1) − η(0), η(1) − η(0))1/2 . Here we have used the following property of the Bochner integral:

Z

Z

b

b

f (t) dt ≤ kf (t)kdt,

a

a valid for any norm on E (see the remarks following Definition 2.2.7). Now let γ : [0, 1] → M be a continuous piecewise C 1 curve, γ(0) = m, γ(1) = n, m ∈ U , where (U, ϕ), ϕ : U → U 0 ⊂ E a chart of M , ϕ(m) = 0. If γ lies entirely in U , then ϕ ◦ γ = η lies entirely in U 0 and the previous estimate gives `(γ) ≥ ag 0 (m0 )(n0 , n0 )1/2 ≥ ar, where r is the radius of the disk U 0 in E about the origin and, n0 = ϕ(n). If γ is not entirely contained in U , then let r be the radius of a disk about the origin and let c ∈ ]0, 1[ be the smallest number for which γ(c) ∩ ϕ−1 ({ x ∈ E | kxk = r }) 6= ∅. Then `(γ) ≥ `(γ|[0, c]) ≥ ag(m0 )((ϕ ◦ γ)(c), (ϕ ◦ γ)(c))1/2 ≥ ar. Thus, we conclude d(m, n) ≥ ar > 0. The equivalence of the original topology of M and of the metric topology defined by d is clear if one notices that they are equivalent in every chart domain U , which in turn is implied by their equivalence in ϕ(U ). 

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377

Notice that the preceding proposition holds in infinite dimensions. 6.5.11 Proposition. A connected Hausdorff n-manifold admits a Riemannian metric if and only if it is second countable. Hence for Hausdorff n-manifolds (not necessarily connected ) paracompactness and metrizability are equivalent. Proof. If M is second countable, it admits partitions of unity by Theorem 6.5.7. Then the patching construction (Theorem 6.5.3) gives a Riemannian metric on M by choosing in every chart the standard inner product in Rn . Conversely, assume M is Riemannian. By Proposition 6.5.10 it is a metric space, which is locally compact and first countable, being locally homeomorphic to Rn . By Theorem 1.6.14, it is second countable.  The main theorem on the existence of partitions of unity in the general case is as follows. 6.5.12 Theorem. Any second-countable or paracompact manifold modeled on a separable Banach space with a C k norm away from the origin admits C k partitions of unity. In particular paracompact (or second countable) manifolds modeled on separable Hilbert spaces admit C ∞ partitions of unity. 6.5.13 Corollary. Paracompact (or second countable) Hausdorff manifolds modeled on separable real Hilbert spaces admit Riemannian metrics. Theorem 6.5.12 will be proved in Supplements 6.5A and 6.5B.1 Partitions of unity are an important technical tool in many proofs. We illustrate this with the sample theorem below which combines differential topological ideas of §3.5, the local and global existence and uniqueness theorem for solutions of vector fields, and partitions of unity. More applications of this sort can be found in the exercises. 6.5.14 Theorem (Ehresmann Fibration Theorem). A proper submersion f : M → N of finite dimensional manifolds with M paracompact is a locally trivial fibration, that is, for any p ∈ N there exists an open neighborhood V of u in N and a diffeomorphism ϕ : V × f −1 (p) → f −1 (V ) such that f (ϕ(x, u)) = x for all x ∈ V and all u ∈ f −1 (p). Proof. Since the statement is local we can replace M , N by chart domains and, in particular, we can assume that N = Rn and p = 0 ∈ Rn . We claim that there are smooth vector fields X1 , . . . , Xn on M such that Xi is f -related to ∂/∂xi ∈ X(Rn ). Indeed, around any point in M such vector fields are easy to obtain using the implicit function theorem (see Theorem 3.5.2). Cover M with such charts, choose a partition of unity subordinate to this covering, and patch these vector fields by means of this partition of unity to obtain X1 , . . . , Xn , f -related to ∂/∂x1 , . . . , ∂/∂xn , respectively. k Let Ft(k) denote the flow of Xk with time variable t(k), k = 1, . . . , n and let t = (t(1), . . . , t(n)) ∈ Rn . If ktk < C, then the integral curves of each Xk starting in f −1 ({ u ∈ Rn | kuk ≤ C }) stay in f −1 { v ∈ Rn | kvk ≤ 2C }, since by Proposition 4.2.4 k (f ◦ Ft(k) )(y) = (f 1 (y), . . . , f k (y) + t(k), . . . , f n (y)).

(6.5.2)

Therefore, since f is proper, Proposition 4.1.19 implies that the vector fields X1 , . . . , Xn are complete. Finally, let ϕ : Rn × f −1 (0) → M be given by 1 n ϕ(t(1), . . . , t(n), u) = (Ft(1) ◦ · · · ◦ Ft(n) )(u)

and note that ϕ is smooth (see Proposition 4.1.17). The map ϕ−1 : M → Rn × f −1 (0) given by n 1 ϕ−1 (m) = (f (m), (F−t(n) ◦ · · · ◦ F−t(1) )(m))

1 There are Hausdorff nonparacompact n-manifolds. These manifolds are necessarily nonmetrizable and do not admit partitions of unity. The standard example of a one-dimensional nonparacompact Hausdorff manifold is the “long line.” In dimensions 2 and 3 such manifolds are constructed from the Pr¨ ufer manifolds. Since nonparacompact manifolds occur rarely in applications, we refer the interested reader to Spivak [1979, Volume 1, Appendix A] for the aforementioned examples.

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is smooth and is easily checked to be the inverse of ϕ. Finally, (f ◦ ϕ)(t, u) = t by equation (6.5.2) since f (u) = 0. 

Supplement 6.5A Partitions of Unity: Reduction to the Local Case We begin with some topological preliminaries. Let S be a paracompact space. If {Uβ } is an open covering of S, it can be refined to a locally finite covering {Wβ }. The first lemma below will show that we can shrink this covering further to get another one, {Vα } such that cl(Vα ) ⊂ Wα with the same indexing set. A technical device used in the proof is the concept of a well-ordered set. An ordered set A in which any two elements can be compared is called well-ordered if every subset has a smallest element (see the introduction to Chapter 1). 6.5.15 Lemma (Shrinking Lemma). Let S be a normal space and let {Wα }α∈A be a locally finite open covering of S. Then there is a locally finite open refinement {Vα }α∈A (with the same indexing set) such that cl(Vα ) ⊂ Wα . Proof. Well-order the indexing set A and call its smallest element α(0). The set C0 defined as C0 = S S\ α>α(0) Wα is closed, so by normality there exists an open set Vα(0) such that C0 ⊂ cl(Vα(0) ) ⊂ Wα(0) . If Vγ is defined for all γ < α, put [ [ Cα = S\{( Vγ ) ∪ ( Wγ )} γ<α

γ>α

and by normality find Vα such that Cα ⊂ cl(Vα ) ⊂ Wα . The collection {Vα }α∈A is the desired locally finite refinement of {Wα }α∈A , provided we can show that it covers S. Given s ∈ S, by local finiteness of the covering {Wα }α∈A , s belongs to only a finite collection of them, say W1 , W2 , . . . , Wn , corresponding to the elements α1 , . . . , αn of the index set. If β denotes the maximum of the elements α1 , . . . , αn , then s 6∈ Wγ for all γ > β, so that if in addition s 6∈ Vγ for all γ < β, then s ∈ Cβ ⊂ Vβ , that is, s ∈ Vβ .  6.5.16 Lemma (A. H. Stone). Proof.

Every pseudometric space is paracompact.

Let {Uα }α∈A be an open covering of the pseudometric space S with distance function d. Put Un,α = { x ∈ Uα | d(x, S\Uα ) ≥ 1/2n }.

By the triangle inequality, we have the inequality d(Un,α , S\Un+1,α ) ≥

1 1 1 − n+1 = n+1 . 2n 2 2

S Well-order the indexing set A and let Vn,α = Un,α \ β<α Un+1,β . If γ, δ ∈ A, we have Vn,γ ⊂ S\Un+1,δ , if γ < δ, and Vn,δ ⊂ S\Un+1,γ if δ < γ. But in both cases we have d(Vn,γ , Vn,δ ) ≥ 1/2n+1 . Define Wn,α = { s ∈ S | d(s, Vn,α ) < 1/2n+3 }, and observe that d(Wn,α , Wn,β ) ≥ 1/2n+2 . Thus for a fixed n, every point s ∈ S has a neighborhood intersecting at most one member of the family { Wn,α | α ∈ A }. Hence { Wn,α | n ∈ N, α ∈ A } is a locally finite open refinement of {Uα }.  Let us now turn to the question of the existence of partitions of unity subordinate to any open covering.

6.5 Partitions of Unity

6.5.17 Proposition (R. Palais). following are equivalent:

379

Let M be a paracompact manifold modeled on the Banach space E. The

(i) M admits C k partitions of unity; (ii) any open covering of M admits a locally finite refinement by C k carriers; (iii) for any open sets O1 , O2 such that cl(O1 ) ⊂ O2 , there exists a C k carrier V such that O1 ⊂ V ⊂ O2 ; (iv) every chart domain of M admits C k partitions of unity subordinate to any open covering; (v) E admits C k partitions of unity subordinate to any open covering of E. Proof. (i) ⇒ (ii). If {(Ui , gi )} is a C k -partition of unity subordinate to an open covering, then clearly carr gi forms a locally finite refinement of the covering by C k carriers. (ii) ⇒ (iii). Let {Vα }α∈A be a locally finite refinement of the open covering {O2 , S\cl(O1 )} by C k carriers and denote by fα ∈ F k (M ), the function for which carr fα = Vα . Let B = { α ∈ A | Vα ⊂ O2 }. Put V =

T

β∈B

Vβ , f = Σβ∈B fβ and notice that O1 ⊂ V ⊂ O2 , carr(f ) = V .

(iii) ⇒ (iv). Let U be any chart domain of M . Then U is diffeomorphic to an open set in E which is a metric space, so is paracompact by Stone’s theorem (Lemma 6.5.16). Let {Uα } be an arbitrary open covering of U and {Vβ } be a locally finite refinement. By the shrinking lemma we may assume that cl(Vβ ) ⊂ U . Again by the shrinking lemma, refine further to a locally finite covering {Wβ } such that cl(Wβ ) ⊂ Vβ . But by (iii) there exists a C k -carrier Oβ such that Wβ ⊂ Oβ ⊂ Vβ , and so {Oβ } is a locally finite refinement of {Uα } by C k -carriers, whose corresponding functions we denote by fβ . Thus f = Σβ fβ is a C k map and {(Vβ , fβ /f )} is a C k partition of unity subordinate to {Uα }. (iv) ⇒ (v). Consider now any open covering {Uα }α∈A of E and let (U, ϕ) be an arbitrary chart of M . Refine first the covering of E by taking the intersections of all its elements with all translates of ϕ(U ). Since E is paracompact, refine again to a locally finite open covering {Vβ }. The inverse images by translations and ϕ of these open sets are subsets of U , hence chart domains, and thus by (iv) they admit partitions of unity subordinate to any covering. Thus every Vβ admits a C k partition of unity subordinate to any open covering, for example to { Vβ ∩ Uα | α ∈ A }; call it {giβ }. Then g = Σi,β giβ is a C k map and the double-indexed set of functions giβ /g forms a C k partition of unity of E. (v) ⇒ (iv). If E admits C k partitions of unity subordinate to any open covering, then so does every open subset by the (already proved) implication (i) ⇒ (ii) applied to M = E, which is paracompact by Theorem 6.5.16. Thus if (U, ϕ) is a chart on M , U admits partitions of unity, since ϕ(U ) does. Finally, we show (iv) implies (i) . Choosing a locally finite atlas, this proof repeats the one given in the last part of (iv) ⇒ (v).  As an application of this proposition we get the following. 6.5.18 Proposition.

Every paracompact n-manifold admits C ∞ -partition of unity.

Proof. By Proposition 6.5.17(ii) and (v) it suffices to show that every open set in Rn is a C ∞ carrier. Any open set U is a countable union of open disks Di . By Lemma 4.2.13, Di = carr (fi ), for some C ∞ function fi : Rn → R. Put Mi = sup{ kDk fi (x)k | x ∈ Rn , k ≤ i } and let f=

∞ X fi . i 2 Mi i=1

By Exercise 2.4-10, f is a C ∞ function for which carr(f ) = U clearly holds.



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In particular, second-countable n-manifolds admit partitions of unity, recovering Theorem 6.5.7.

Supplement 6.5B Partitions of Unity: The Local Case Proposition 6.5.17 reduces the problem of the existence of partitions of unity to the local one, namely finding partitions of unity in Banach spaces. This problem has been studied by Bonic and Frampton [1966] for separable Banach spaces. 6.5.19 Proposition ( Bonic and Frampton [1966]). equivalent.

Let E be a separable Banach space. The following are

(i) Any open set of E is a C k carrier. (ii) E admits C k partitions of unity subordinate to any open covering of E. (iii) There exists a bounded nonempty C k carrier in E. Proof. By Proposition 6.5.17, (i) and (ii) are equivalent since E is paracompact by Lemma 6.5.16. It remains to be shown that (iii) implies (i), since clearly (ii) implies (iii). This proceeds in several steps. First, we show that any neighborhood contains a C k carrier. Let U be any open set and let carr(f ) ⊂ Dr (0) be the bounded carrier given by (iii), f ∈ C k (E), f ≥ 0. Let e ∈ U , fix e0 ∈ carr(f ), and choose ε > 0 such that Dε (e) ⊂ U . Define g ∈ C k (E), g ≥ 0 by g(v) = f (K(v − e) + e0 ),

K > 0,

where K remains to be determined from the condition that carr(g) ⊂ Dε (e). An easy computation shows that if K > (r + ke0 k)/ε, this inclusion is verified. Since e ∈ carr(g), carr(g) is an open neighborhood of e. Second, we show that any open set can be covered by a countable locally finite family of C k carriers. k By the S first step, the openk set U can be covered by a family of C carriers. By Lindel¨of’s lemma 1.1.6, U = n Vn where Vn is a C carrier, the union being over the positive integers. We need to find a refinement of this covering by C k carriers. Let fn ∈ C k (E) be such that carr fn = Vn . Define Un = { e ∈ E | fn (e) > 0, fi (e) < 1/n for all i < n }. Clearly U1 = V1 and inductively " Un = Vn ∩

# \

fi−1 (]−∞, 1/n[)

.

i
By the composite mapping theorem, the inverse image of a C k carrier is a C k carrier, so that fi−1 (]−∞, 1/n[) is a C k carrier, since ]∞, 1/n[ is a C k carrier in R (see the proof of Proposition 6.5.18). Finite intersections of C k carriers is a C k carrier (just take the product of the functions in question) so that Un is also a C k carrier. Clearly Un ⊂ Vn . We shall prove that {Un } is a locally finite open covering of U . Let e ∈ U . If e ∈ Vn for all n, then clearly e ∈ U1 = V1 . If not, then there exists a smallest n, say N , such that e ∈ VN . Then fi (e) = 0 for i < N and thus e ∈ UN = { e ∈ E | fN (e) > 0, fi (e) < 1/N for all i < N }. Thus, the sets Un cover U . This open covering is also locally finite for if e ∈ Vn and N is such that f (e) > 1/N , then the neighborhood { u ∈ U | fn (e) > 1/N } has empty intersections with all US m for m > N . Third we show that the open set U is a C k -carrier. By the second step, U = n Un , with Un a locally finite open covering of U by C k carriers. Then f = Σn fn is C k , f (e) ≥ 0 for all e ∈ E and carr(f ) = U . 

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381

The separability assumption was used only in showing that (iii) implies (i). There is no general theorem known to us for nonseparable Banach spaces. Also, it is not known in general whether Banach spaces admit bounded C k carriers, for k ≥ 1. However, we have the following. 6.5.20 Proposition. bounded C k -carriers.

If the Banach space E has a norm C k away from its origin, k ≥ 1, then E has

Proof. By Corollary 4.2.14 there exists ϕ : R → R, C ∞ with compact support and equal to one in a neighborhood of the origin. If k · k : E\{0} → R is C k , k ≥ 1, then ϕ ◦ k · k : E\{0} → R is a nonzero map which is C k , has bounded support k · k−1 (supp ϕ), and can be extended in a C k manner to E.  Theorem 6.5.12 now follows from Propositions 6.5.20, 6.5.19, and 6.5.17. The situation with regard to Banach subspaces and submanifolds is clarified in the following proposition, whose proof is an immediate consequence of Propositions 6.5.19 and 6.5.17. 6.5.21 Proposition. (i) If E is a Banach space admitting C k partitions of unity then so does any closed subspace, (ii) If a manifold admits C k partitions of unity subordinate to any open covering, then so does any submanifold. We shall not develop this discussion of partitions of unity on Banach manifolds any further, but we shall end by quoting a few theorems that show how intimately connected partitions of unity are with the topology of the model space. By Propositions 6.5.19 and 6.5.20, for separable Banach spaces one is interested whether the norm is C k away from the origin. Restrepo [1964] has shown that a separable Banach space has a C 1 norm away from the origin if and only if its dual is separable. Bonic and Reis [1966] and Sundaresan [1967] have shown that if the norms on E and E∗ are differentiable on E\{0} and E∗ \{0}, respectively, then E is reflexive, for E a real Banach space (not necessarily separable). Moreover, E is a Hilbert space if and only if the norms on E and E∗ are twice differentiable away from the origin. This result has been strengthened by Leonard and Sunderesan [1973], who show that a real Banach space is isometric to a Hilbert space if and only if the norm is C 2 away from the origin and the second derivative of e 7→ kek2 /2 is bounded by 1 on the unit sphere; see Rao [1972] for a related result. Palais [1965b] has shown that any paracompact Banach manifold admits Lipschitz partitions of unity. Because of the importance of the differentiability class of the norm in Banach spaces there has been considerable work in the direction of determining the exact differentiability class of concrete function spaces. Thus Bonic and Frampton [1966] have shown that the canonical norms on the spaces Lp (R), `p (R), p ≥ 1, p < ∞ are C ∞ away from the origin if p is even, C p−1 with D(k · kp−1 ) Lipschitz, if p is odd, and C [p] with D[p] (k · kp ) H¨ older continuous of order p − [p], if p is not an integer. The space c0 of sequences of real numbers convergent to zero has an equivalent norm that is C ∞ away from the origin, a result due to Kuiper. Using this result, Frampton and Tromba [1972] show that the Λ-spaces (closures of C ∞ in the H¨older norm) admit a C ∞ norm away from the origin. The standard norm on the Banach space of continuous real valued functions on [0, 1] is nowhere differentiable. Moreover, since C 0 ([0, 1], R) is separable with nonseparable dual, it is impossible to find an equivalent norm that is differentiable away from the origin. To our knowledge it is still an open problem whether C 0 ([0, 1], R) admits C ∞ partitions of unity for k ≥ 1. Finally, the only results known to us for nonseparable Hilbert spaces are those of Wells [1971, 1973], who has proved that nonseparable Hilbert space admits C 2 partitions of unity. The techniques used in the proof, however, do not seem to indicate a general way to approach this problem.

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Supplement 6.5C Simple Connectivity of Fiber Bundles The goal of this supplement is to discuss the homotopy lifting property for locally trivial continuous fiber bundles over a paracompact base. This theorem is shown to imply an important criterion on the simple connectedness of the total space of fiber bundles with paracompact base. 6.5.22 Theorem (Homotopy Lifting Theorem). Let π : E → B be a locally trivial C 0 fiber bundle and let M be a paracompact topological space. If h : [0, 1] × M → B is a continuous homotopy and f : M → E is any continuous map satisfying π ◦ f = h(0, ·), there exists a continuous homotopy H : [0, 1] × M → E satisfying π ◦ H = h and H(0, ·) = f . If in addition, h fixes some point m ∈ M , that is h(t, m) is constant for t in a segment ∆ of [0, 1], then H(t, m) is also constant for t ∈ ∆. Remark. The property in the statement of the theorem is called the homotopy lifting property . A Hurewicz fibration is a continuous surjective map π : E → B satisfying the homotopy lifting property relative to any topological space M . Thus the theorem above says that a locally trivial C 0 fiber bundle is a Hurewicz fibration relative to paracompact spaces.  See Steenrod [1957] and Huebsch [1955] for the proof. 6.5.23 Corollary. Let π : E → B be a C 0 locally trivial fiber bundle. If the base B and the fiber F are simply connected, then E is simply connected. Proof. Let c : [0, 1] → E be a loop, c(0) = c(1) = e0 . Then d = π ◦ c is a loop in B based at π(e0 ) = b0 . Since B is simply connected there is a homotopy h : [0, 1] × [0, 1] → B such that h(0, t) = d(t), h(1, t) = b0 for all t ∈ [0, 1], and h(s, 0) = h(s, 1) = b0 for all s ∈ [0, 1]. By the homotopy lifting theorem there is a homotopy H : [0, 1] × [0, 1] → E such that π ◦ H = h, H(0, ·) = c, and H(s, 0) = H(0, 0) = c(0) = e0 ,

H(s, 1) = c(1) = e0 .

Since (π ◦ H)(1, t) = h(1, t) = b0 , it follows that t 7→ H(1, t) is a path in π −1 (b0 ) starting at H(1, 0) = e0 and ending also at H(1, 1) = e0 . Since π −1 (b0 ) is simply connected, there is a homotopy k : [1, 2]×[0, 1] → π −1 (b0 ) such that k(1, t) = H(1, t), k(2, t) = e0 for all t ∈ [0, 1] and k(s, 0) = k(s, 1) = e0 for all s ∈ [1, 2]. Define the continuous homotopy K : [0, 2] × [0, 1] → E by ( H(s, t), if s ∈ [0, 1]; K(s, t) = k(s, t), if s ∈ [1, 2], and note that K(0, t) = H(0, t) = c(t),

K(2, t) = k(2, t) = e0

for any t ∈ [0, 1], and K(s, 1) = e0 for any s ∈ [0, 2]. Thus, c is contractible to e0 and E is therefore simply connected. 

6.5 Partitions of Unity

383

Exercises  6.5-1 (Whitney). Show that any closed set F in Rn is the inverse image of 0 by a C ∞ real-valued positive function on Rn . Generalize this to any n-manifold. Hint: Cover Rn \F with a sequence of open disks Dn and choose for each n a smooth function χn ≥ 0, satisfying χn |Dn > 0, with the absolute value of χn and all its derivatives ≤ 2n . Set χ = Σn≥0 χn .  6.5-2.

In a paracompact topological space, an open subset need not be paracompact. Prove the following.

(i) If every open subset of a paracompact space is paracompact, then any subspace is paracompact. (ii) Every open submanifold of a paracompact manifold is paracompact. Hint: Use chart domains to conclude metrizability.  6.5-3. Let π : E → M be a vector bundle, E 0 ⊂ E a subbundle and assume M admits C k partitions of unity subordinate to any open covering. Show that E 0 splits in E, that is, there exists a subbundle E 00 such that E = E 00 ⊕ E 00 . Hint: The result is trivial for local bundles. Construct for every element of a locally finite covering {Ui } a vector bundle map fi whose kernel is the complement of E 0 |Ui . For {(Ui , gi )} a C k partition of unity, put f = Σi gi fi and show that E = E 0 ⊕ ker f .  6.5-4. Let π : E → M be a vector bundle over the base M that admits C k partitions of unity and with the fibers of E modeled on a Hilbert space. Show that E admits a C k bundle metric, that is, a C k map g : M → T20 (E) that is symmetric, strongly nondegenerate, and positive definite at every point m ∈ M .  6.5-5. Let E → M be a line bundle over the manifold M admitting C k partitions of unity sub-ordinate to any open covering. Show that E × E is trivial Hint: E × E = L(E ∗ , E) and construct a local base that can be extended.  6.5-6. Assume M admits C k partitions of unity. Show that any submanifold of M diffeomorphic to S 1 is the integral curve of a C k vector field on M .  6.5-7. Let M be a connected paracompact manifold. Show that there exists a C ∞ proper mapping f : M → Rk . Hint: M is second countable, being Riemannian. Show the statement for k = 1, where f = Σi≥1 iϕi , and {ϕi } is a countable partition of unity.  6.5-8. Let M be a connected paracompact n-manifold and X ∈ X(M ). Show that there exist h ∈ F(M ), h > 0 such that Y = hX is complete. Hint: With f as in Exercise 6.5-7 put h = exp{−(X[f ])2 } so that |Y [f ]| ≤ 1. Hence (f ◦c)(]a, b[) is bounded for any integral curve c of Y and ]a, b[ in the domain of c.  6.5-9.

Let M be a paracompact, non-compact manifold.

(i) Show that there exists a locally finite sequence of open sets { Ui | i ∈ Z } such that Ui ∩ Ui+1 6= ∅ unless j = i − 1, i, i + 1, and each Ui is a chart domain diffeomorphic (by the chart map ϕi ) with the open unit ball in the model space of M . See Figure 6.5.1. Hint: Let V be a locally finite open cover of M with chart domains diffeomorphic by their chart maps to the open unit ball such that no finite subcover of V covers M , and no two elements of V include each other. Let U0 , U1 be distinct elements of V, U0 ∩ U1 6= ∅. Let U−1 ∈ V be such that U−1 ∩ U0 6= ∅; U−1 ∩ U1 = ∅; such a U−1 exists by local finiteness of V. Now use induction. (ii) Use (i) to show that there exists an embedding of R in M as a closed manifold. Hint: Let c0 be a smooth curve in U0 diffeomorphic by ϕ0 to ]−1, 1[ connecting a point in U−1 ∩ U0 to a point in U0 ∩ U1 . Next, extend c0 ∩ U1 smoothly to a curve c1 , diffeomorphic by ϕ1 to ]0, 2[ ending inside U1 ∩ U2 ; show that c1 ∩ U0 extends the curve c0 ∩ U1 inside U0 ∩ U1 . Now use induction.

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Figure 6.5.1. A chart chain

(iii) Show that on each non-compact paracompact manifold admitting partitions of unity there exists a non-complete vector field. Hint: Embed R in M by (ii) and on R consider the vector field x˙ = x2 . Extend it to M via partitions of unity.  6.5-10. Show that every compact n-manifold embeds in some Rk for k big enough in the following way. If {(Ui , ϕi )}i=1,...,N is a finite atlas with ϕi (Ui ) the ball of radius 2 in Rn , let χ ∈ C ∞ (Rn ), χ = 1 on the ball of radius 1 and χ = 0 outside the ball of radius 2. Put fi = (χ ◦ ϕi ) · ϕi : M → Rn , where fi = 0 outside Ui . Show that fi is C ∞ and that ψ : M → RN n × RN , defined by ψ(m) = (f1 (m), . . . , fN (m), χ(ϕ1 (m)), . . . , χ(ϕN (m))) is an embedding.  6.5-11.

Let g be a Riemannian metric on M .

(i) Show that if N is a submanifold, its g-normal bundle νg (N ) = { v ∈ Tn M | n ∈ N, v ⊥ Tn N } is a subbundle of T M . (ii) Show that T M |N = νg (N ) ⊕ T N . (iii) If h is another Riemannian metric on M , show that νg (N ) is a vector bundle isomorphic to νh (N ).  6.5-12. Show that if f : M → N is a proper surjective submersion with M paracompact and N connected, then it is a locally trivial fiber bundle. Hint: To show that all fibers are diffeomorphic, connect a fixed point of N with any other point by a smooth path and cover the path with the neighborhoods in N given by Theorem 6.5.14.

Page 385

7 Differential Forms

This version: February 15, 2007

Differential k-forms are tensor fields of type (0, k) that are completely antisymmetric. Such tensor fields arise in many applications in physics, engineering, and mathematics. A hint at why this is so is the fact that the classical operations of grad, div, and curl and the theorems of Green, Gauss, and Stokes can all be expressed concisely in terms of differential forms and an operator on differential forms to be studied in this chapter, the exterior derivative d. One of the most compelling facts is that the standard vector identities ∇×(∇f ) = 0 and ∇·(∇×X) = 0 are elegantly phrased as the single identity d2 = 0. However, the examples of Hamiltonian mechanics and Maxwell’s equations (see Chapter 9) show that their applicability goes well beyond this. The goal of the chapter is to develop the calculus of differential forms, due largely to Cartan [1945]. The exterior derivative operator d plays a central role; its properties and the expression of the Lie derivative in terms of it will be developed.

7.1

Exterior Algebra

We begin with the exterior algebra of a vector space and extend this fiberwise to a vector bundle. As with tensor fields, the most important case is the tangent bundle of a manifold, which is considered in the next section. We first recall a few facts about the permutation group on k elements; some of these facts have already been discussed in §2.2 (see the discussion before Definition 2.2.10). Proofs of the results that we cite are obtainable from virtually any elementary algebra book. The permutation group on k elements, denoted Sk , consists of all bijections σ : {1, . . . , k} → {1, . . . , k} usually given in the form of a table   1 ··· k , σ(1) · · · σ(k) together with the structure of a group under composition of maps. Clearly, Sk has order k!. Letting {−1, 1} have its natural multiplicative group structure, there is a homomorphism denoted sign : Sk → {−1, 1}; that is, for σ, τ ∈ Sk , sign(σ ◦ τ ) = (sign σ)(sign τ ). A permutation σ is called even when sign σ = +1 and odd when sign σ = −1. This homomorphism can be described as follows. A transposition is a permutation that

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7. Differential Forms

swaps two elements of {1, . . . , k}, leaving the remainder fixed. An even (odd) permutation can be written as the product of an even (odd) number of transpositions. The expression of σ as a product of transpositions is not unique, but the number of transpositions is always even or odd corresponding to σ being even or odd. If E and F are Banach spaces, an element of Tk0 (E, F) = Lk (E; F); that is, a k-multilinear continuous mapping of E × · · · × E → F is called skew symmetric when t(e1 , . . . , ek ) = (sign σ) t(eσ(1) , . . . , eσ(k) ) for all e1 , . . . , ek ∈ E and σ ∈ Sk . This is equivalent to saying that t(e1 , . . . , ek ) changes sign when any two of e1 , . . . , ek are swapped. The subspace of skew symmetric elements of Lk (E; F) is denoted Lka (E; F) (the V0 subscript a stands for “alternating”). Some additional shorthand will be useful. Namely, let (E, F) = F, V1 Vk (E, F) = L(E, F) and in general, (E, F) = Lka (E; F), be the vector space of skew symmetric F-valued V0 V1 multilinear maps or exterior F-valued k-forms on E. If F = R, we write (E) = R, (E) = E∗ and Vk V k (E) = Lka (E; R); elements of (E) are called exterior k-forms.1 Vk We can form elements of (E, F) by skew symmetrizing elements of Tk0 (E; F). For example, if t ∈ T20 (E), the two tensor At defined by 1 (At) (e1 , e2 ) = [t (e1 , e2 ) − t (e2 , e1 )] 2 is skew symmetric and if t is already skew, At coincides with t. More generally, we make the following definition. 7.1.1 Definition. The alternation mapping A : Tk0 (E, F) → Tk0 (E, F) (for notational simplicity we do not index the A with E, F or k) is defined by At(e1 , . . . , ek ) =

1 X (sign σ) t(eσ(1) , . . . , eσ(k) ), k! σ∈Sk

where the sum is over all k! elements of Sk .

Proof.

Vk

Vk (E, F), A| (E, F) is the identity, and A ◦ A = A. Vk Linearity of A is clear from the definition. If t ∈ (E, F), then

7.1.2 Proposition.

A is a linear mapping onto

1 X (sign σ) t(eσ(1) , . . . , eσ(k) ) k! σ∈Sk 1 X = t(e1 , . . . , ek ) = t(e1 , . . . , ek ) k!

At(e1 , . . . , ek ) =

σ∈Sk

since Sk has order k!. This proves the first two assertions, and the last follows from them.



From A = A ◦ A, it follows that kAk ≤ kAk2 , and so, as A 6= 0, kAk ≥ 1. From the definition of A, we see kAk ≤ 1; thus kAk = 1. In particular, A is continuous. Vk+l 7.1.3 Definition. If α ∈ Tk0 (E) and β ∈ Tl0 (E), define their wedge product α ∧ β ∈ (E) by α∧β =

(k + l)! A(α ⊗ β). k!l!

For F-valued forms, we can also define ∧, where ⊗ is taken with respect to a given bilinear form B ∈ L(F1 , F2 ; F3 ). Since A and ⊗ are continuous, so is ∧. There are several possible conventions for defining the wedge product ∧. The one here conforms to Spivak [1979], and Bourbaki [1971] but not to Kobayashi 1 Note,

however, that some authors write

Vk

(E∗ ) where we write

Vk

(E).

7.1 Exterior Algebra

387

and Nomizu [1963] or Guillemin and Pollack [1974]. See Exercise 7.1-7 for the possible conventions. Our definition of α ∧ β is the one that eliminates the largest number of constants encountered later. A (k, l)-shuffle is a permutation σ of {1, 2, . . . , k + l} such that σ(1) < · · · < σ(k)

and σ(k + 1) < · · · < σ(k + l).

The reason for the name “shuffles” is that these are the kind of permutations made when a deck of k + l cards is shuffled, with k cards held in one hand and l in the other. The reader should prove that for α a k-form and β an l-form, we have X (α ∧ β)(e1 , . . . , ek+l ) = (sign σ)α(eσ(1) , . . . , eσ(k) )β(eσ(k+1) , . . . , eσ(k+l) ), (7.1.1) where the sum is over all (k, l) shuffles σ. Formula (7.1.1) is a convenient way to compute wedge products, as we see in the following examples. 7.1.4 Examples. A.

If α is a two-form and β is a one-form, then (α ∧ β)(e1 , e2 , e3 ) = α(e1 , e2 )β(e3 ) − α(e1 , e3 )β(e2 ) + α(e2 , e3 )β(e1 )

Indeed the only (2, 1) shuffles in S3 are   1 2 3 , 1 2 3



1 1

2 3

3 2



 ,

and

1 2

2 3

3 1

 ,

of which only the second one has sign −1. B.

If α and β are one-forms, then (α ∧ β)(e1 , e2 ) = α(e1 )β(e2 ) − α(e2 )β(e1 )

since S2 consists of two (1, 1) shuffles. 7.1.5 Proposition.



0 (E), we have For α ∈ Tk0 (E), β ∈ Tl0 (E), and γ ∈ Tm

(i) α ∧ β = Aα ∧ β = α ∧ Aβ; (ii) ∧ is bilinear ; (iii) α ∧ β = (−1)kl β ∧ α; (iv) α ∧ (β ∧ γ) = (α ∧ β) ∧ γ = Proof.

(k + l + m)! A(α ⊗ β ⊗ γ). k!l!m!

For (i), first note that if σ ∈ Sk and we define
σt (e1 , . . . , ek ) = t eσ(1) , . . . , eσ(k) ,

then A(σt) = (sign σ)At, because A(σt)(e1 , . . . , ek ) =

1 X (sign ρ) t(eρσ(1) , . . . , eρσ(k) ) k! ρ∈Sk

1 X = (sign σ)(sign ρσ) t(eρσ(1) , . . . , eρσ(k) ) k! ρ∈Sk

= (sign σ)At(e1 , . . . , ek )

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since ρ 7→ ρσ is a bijection. Therefore, since At =

1 X (sign σ)σt, k! σ∈Sk

and using linearity of A, we get ! 1 X A(Aα ⊗ β) = A (sign τ )(τ α ⊗ β) k! τ ∈Sk 1 X = (sign τ )A(τ α ⊗ β) k! τ ∈Sk 1 X (sign τ 0 )Aτ 0 (α ⊗ β), = k! τ ∈Sk

where τ 0 ∈ Sk+l is defined by τ 0 (1, . . . , k, . . . , k + l) = (τ (1), . . . , τ (k), k + 1, . . . , k + l), so sign τ = sign τ 0 and τ α ⊗ β = τ 0 (α ⊗ β). Thus the preceding expression for A(Aα ⊗ β) becomes 1 X 1 X (sign τ 0 )(sign τ 0 )A(α ⊗ β) = A(α ⊗ β) 1 = A(α ⊗ β). k! k! τ ∈Sk

τ ∈Sk

Thus, A(Aα ⊗ β) = A(α ⊗ β) which is equivalent to (Aα) ∧ β = α ∧ β. The other equality in (i) is similar. (ii) is clear since ⊗ is bilinear and A is linear. For (iii), let σ0 ∈ Sk+l be given by σ0 (1, . . . , k + l) = (k + 1, . . . ., k + l, 1, . . . , k) . Then
(α ⊗ β) (e1 , . . . , ek+l ) = (β ⊗ α) eσ0 (1) , . . . , eσ0 (k+l) . Hence, by the proof of (i), A(α ⊗ β) = (sign σ0 )A(β ⊗ α). But sign σ0 = (−1)kl . Finally, for (iv), (k + l + m)! A(α ⊗ (β ∧ γ)) k!(l + m)! (k + l + m)! (l + m)! = A(α ⊗ A(β ⊗ γ)) k!(l + m)! l!m! (k + l + m)! = A(α ⊗ β ⊗ γ) k!l!m!

α ∧ (β ∧ γ) =

since A(α ⊗ Aβ) = A(α ⊗ β), which was proved in (i), and by associativity of ⊗. We calculate (α ∧ β) ∧ γ in the same way.  Conclusions (i)–(iii) hold (with identical proofs) for F-valued forms when the wedge product is taken with respect to a given bilinear mapping B. Associativity can also be generalized under suitable assumptions on the bilinear mappings, such as requiring F to be an associative algebra under B. Because of associativity, α ∧ β ∧ γ can be written with no ambiguity.

7.1 Exterior Algebra

389

7.1.6 Examples. A.

If αi , i = 1, . . . , k are one-forms, then X


α1 ∧ · · · ∧ αk (e1 , . . . , ek ) = (sign σ) α1 eσ(1) · · · αk eσ(k) σ

  = det αi (ej ) . Indeed, repeated application of Proposition 7.1.5(iv) gives γ1 ∧ · · · ∧ γk =

(d1 + · · · + dk )! A(γ 1 ⊗ · · · ⊗ γ k ), d1 ! · · · dk !

(7.1.2)

where γi is a di -form on E. In particular, if αi is a one form, equation 7.1.2 gives α1 ∧ · · · ∧ αk = k!A(α1 ⊗ · · · ⊗ αk ),

(7.1.3)

which yields the stated formula after using the definition of A. If e1 , . . . , en and e1 , . . . , en are dual bases, observe that as a special case,
e1 ∧ · · · ∧ ek (e1 , . . . , ek ) = 1.

B. If at least one of α or β is of even degree, then Proposition 7.1.5(iii) says that α ∧ β = β ∧ α. If both are of odd degree, then α ∧ β = −β ∧ α. Thus, if α is a one-form, if α is a two-form, V2then4
α ∧ α = 10. But then in general α ∧ α 6= 0. For example, if α = e1 ∧ e2 + e3 ∧ e4 ∈ R where e , e2 , e3 , e4 is the standard dual basis of R4 , then α ∧ α = 2e1 ∧ e2 ∧ e3 ∧ e4 6= 0. C. The properties listed in Proposition 7.1.5 make the computations of wedge products similar to polynomial multiplication, care being taken with commutativity. For example, if α1 , . . . , α5 are one forms on R5 , V2 5
α = 2α1 ∧ α3 + α2 ∧ α3 − 3α3 ∧ α4 ∈ R and β = −α1 ∧ α2 ∧ α5 + 2α1 ∧ α3 ∧ α4 ∈

V3


R5 ,

then the wedge product α ∧ β is computed using the bilinearity and commutation properties of ∧: α ∧ β = −2(α1 ∧ α3 ) ∧ (α1 ∧ α2 ∧ α5 ) − (α2 ∧ α3 ) ∧ (α1 ∧ α2 ∧ α5 ) + 3(α3 ∧ α4 ) ∧ (α1 ∧ α2 ∧ α5 ) + 4(α1 ∧ α3 ) ∧ (α1 ∧ α3 ∧ α4 ) + 2(α2 ∧ α3 ) ∧ (α1 ∧ α3 ∧ α4 ) − 6(α3 ∧ α4 ) ∧ (α1 ∧ α3 ∧ α4 ) = 3α3 ∧ α4 ∧ α1 ∧ α2 ∧ α5 = 3α1 ∧ α2 ∧ α3 ∧ α4 ∧ α5 .



To express the wedge product in coordinate notation, suppose E is finite dimensional with basis e1 , . . . , en . The components of t ∈ Tk0 (E) are the real numbers ti1 ···ik = t(ei1 , . . . , eik ),

1 ≤ i1 , . . . , ik ≤ n.

(7.1.4)

Vk V2 For t ∈ (E), equation (7.1.4) is antisymmetric in its indices i1 , . . . , ik . For example, t ∈ (E) yields tij , a skew symmetric n × n matrix. From Definition 7.1.1 of the alternation mapping and equation 7.1.4, we have 1 X (sign σ)tσ(i1 )···σ(ik ) ; (At)i1 ...ik = k! σ∈Sk

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that is, At antisymmetrizes the components of t. For example, if t ∈ T20 (E), then (At)ij = If α ∈

Vk

(E) and β ∈

Vl

tij − tji . 2

(E), then equations (7.1.1) and (7.1.4) yield X (α ∧ β)i1 ···ik+l = (sign σ)ασ(i1 )...σ(ik ) βσ(ik+1 )...σ(ik+l )

where the sum is over all the (k, l)-shuffles in Sk+l . Vk 7.1.7 Definition. The direct sum of the spaces (E) (k = 0, 1, 2, . . . ) together with its structure of real vector space and multiplicationVinduced by ∧, is called the exterior algebra of E, or the Grassmann algebra of E. It is denoted by (E). V Thus (E) is a graded associative algebra, that is, an algebra in which every element has a degree (a k-form has degree k), and the degree map is additive on products (by Proposition 7.1.2 and Definition 7.1.3). V Elements of (E) may be written as finite sums of increasing degree exactly as one writes a polynomial as V1 V2 V a sum of monomials. Thus if a, b, c ∈ R, α ∈ (E) and β ∈ (E) then a + bα + cβ makes sense in (E). V1 V The one-form α can be understood as an element of (E) and also of (E), where α is identified with 0 + α + 0 + 0 + ··· . Vk 7.1.8 Proposition. Suppose E is finite dimensional and n = dim E. Then for k > n, (E) = {0}, Vk while for 0 < k ≤ n, (E) has dimension n!/(n − k)!k!. The exterior algebra over E has dimension 2n . If Vk {e1 , . . . , en } is an (ordered ) basis of E and {e1 , . . . , en } its dual basis, a basis of (E) is  i1 e ∧ · · · ∧ eik | 1 ≤ i1 < i2 < · · · < ik ≤ n . (7.1.5) Proof. 6.1.2,

First we show that the indicated wedge products span

Vk

(E). If α ∈

Vk

(E), then from Proposition

α = α(ei1 , . . . , eik )ei1 ⊗ · · · ⊗ eik , where the summation convention indicates that this should be summed over all choices of i1 , . . . , ik between 1 and n. If the linear operator A is applied to this sum and equation (7.1.2) is used, we get α = Aα = α(ei1 , . . . , eik )A(ei1 ⊗ · · · ⊗ eik ) 1 = α(ei1 , . . . , eik ) ei1 ∧ · · · ∧ eik . k! The sum still runs over all choices of the i1 , . . . , ik and we want only distinct, ordered ones. However, since α is skew symmetric, the coefficient in α is 0 if i1 , . . . , ik are not distinct. If they are distinct and σ ∈ Sk , then α(ei1 , . . . , eik )ei1 ∧ · · · ∧ eik = α(eσ(i1 ) , . . . , eσ(ik ) )eσ(i1 ) ∧ · · · ∧ eσ(ik ) , since both α and the wedge product change by a factor of sign σ. Since there are k! of these rearrangements, we are left with X α(ei1 , . . . , eik )ei1 ∧ · · · ∧ eik . α= i1 <···
Vk This shows that equation (7.1.5) spans (E). Secondly, we show that the elements in equation (7.1.5) are linearly independent. Suppose that X αi1 ,...,ik ei1 ∧ · · · ∧ eik = 0. i1 <···
7.1 Exterior Algebra

391

0 0 For fixed i01 , . . . , i0k , let jk+1 , . . . , jn0 denote the complementary set of indices, jk+1 < · · · < jn0 . Then X 0 0 αi1 ···ik ei1 ∧ · · · ∧ eik ∧ ejk+1 ∧ · · · ∧ ejn = 0. i1 <···
However, this reduces to αi01 ···i0k e1 ∧ · · · ∧ en = 0. But e1 ∧ · · · ∧ en 6= 0, as (e1 ∧ · · · ∧ en )(e1 , . . . , en ) = 1 by Example 7.1.6A. Hence the coefficients are zero.  Vn 7.1.9 V Corollary. If dim E = n, then dim (E) = 1. If {α1 , . . . , αn } is a basis for E∗ , then α1 ∧ · · · ∧ αn n spans (E). Proof.

This follows from Proposition 7.1.8.

7.1.10 Corollary. Vk (E). Proof.



Let α1 , . . . , αk ∈ E∗ . Then α1 , . . . , αk are linearly independent iff α1 ∧ · · · ∧ αk spans

If α1 , . . . , αk are linearly dependent, then αi =

X

cj αj

j6=i

for some i. Since α ∧ α = 0, for α a one-form, we see that α1 ∧ · · · ∧ αk = 0. Conversely, if α1 ∧ · · · ∧ αk = 0, then by Corollary 7.1.9, α1 , . . . , αk is not a basis for span {α1 , . . . , αk }. Therefore k > dim(span{α1 , . . . , αk }) and so α1 , . . . , αk are linearly dependent.  V1 Vk Vk−1 7.1.11 Corollary. Let θ ∈ (E) and α ∈ (E). Then θ ∧ α = 0 iff there exists β ∈ (E) such that α = θ ∧ β. Proof. Clearly, if α = θ ∧ β, then θ ∧ α = 0. Conversely, assume θ ∧ α = 0, θ 6= 0 and choose a basis {ei }i∈I of E such that for some k ∈ I, ek = θ. If X α= αi1 ...ik ei1 ∧ · · · ∧ eik , i1 <···
then θ ∧ α = 0 implies that all summands not involving ek are zero. Now factor ek out of the remaining terms and call the resulting (k − 1)-form β.  7.1.12 Examples. V1 2 (R ) A. Let E = R2 , {e1 , e2 } be the standard basis of R2 and {e1 , e2 } the dual basis. Any element ω of V 2 can be written uniquely as ω = ω1 e1 + ω2 e2 , and any element ω of (R2 ) can be written uniquely as ω = ω12 e1 ∧ e2 . V1 3 B. Let E = R3 , {e1 , e2 , e3 } be the standard basis, and {e1 , e2 , e3 } the dual basis. Any element ω ∈ (R ) can be written uniquely as ω = ω1 e1 + ω2 e2 + ω3 e3 . V2 3 V3 3 Similarly, any elements η ∈ (R ) and ξ ∈ (R ) can be uniquely written as η = η12 e1 ∧ e2 + η13 e1 ∧ e3 + η23 e2 ∧ e3 and ξ = ξ123 e1 ∧ e2 ∧ e3 .

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7. Differential Forms

V1 3
V2 3
R , and C. Since R3 , R all have the same dimension, they are isomorphic. An isomorphism

V ∗ 1 R3 = R3 is the standard one associated to a given basis: ei 7→ ei , i = 1, 2, 3. An isomorphism R3 ∼ = V1 3
V2 3
of R with R is determined by e1 7→ e2 ∧ e3 ,

e2 7→ e3 ∧ e1 , and e3 7→ e1 ∧ e2 . V1 3
V2 3
This isomorphism is usually denoted by ∗ : R 7→ R ; we shall study this map in general in the next section under the name Hodge star V operator . 1 The standard isomorphism of R3 with (R3 ) = (R3 )∗ is given by the index lowering action [ of the V2 3 3 [ i standard metric on R ; that is, (ei ) = e . Then ∗ ◦ [ : R3 → (R ) has the following property:   (7.1.6) ∗ ◦ [ (e × f ) = [(e) ∧ [(f ) for all v, w ∈ R3 , where × denotes the usual cross-product of vectors; that is,


v × w = v 2 w3 − v 3 w2 e1 + v 3 w1 − v 1 w3 e2 + v 1 w2 − v 2 w1 e3 . The relation (7.1.6) follows from the definitions and the fact that if α = α1 e1 + α2 e2 + α3 e3 and β = β1 e1 + β2 e2 + β3 e3 , then α ∧ β = (α2 β3 − α3 β2 ) e2 ∧ e3 + (α3 β1 − α1 β3 ) e3 ∧ e1 + (α1 β2 − α2 β1 ) e1 ∧ e2 .



Exercises  7.1-1. Compute α∧α, α∧β, β ∧β, and β ∧α∧β for α = 2e1 ∧e3 −e2 ∧e3 ∈
∗ where {e1 , e2 , e3 } is a basis of R3 .

V2


R3 and β = −e1 +e2 −2e3

 7.1-2.

If k! is omitted in the definition of A in Definition 7.1.1, show that ∧ fails to be associative. Vr  7.1-3. Let v1 , . . . , vk be linearly dependent vectors. Show that for each α ∈ (E), we have α (v1 , . . . , vk ) = 0. V ∗ Vk ∗ k  7.1-4. Let E be finite dimensional. Show that (E ) is isomorphic to (E) . V ∗ Vk ∗ k Hint: Define ϕ : (E) → (E ) by

ϕ (σ) α1 , . . . , αk = σ α1 ∧ · · · ∧ αk and construct its inverse using the basis in Proposition 7.1.8.  7.1-5. Let {e1 , . . . , en } be a basis of E with dual basis {e1 , . . . , en } and let {f1 , . . . , fm } be a basis of F. Show the following: Vk Vk (E, F) can be uniquely written as β = Σ1≤i≤m βi fi for βi ∈ (E), where (i) every β ∈ (γf ) (v1 , . . . , vk ) = γ (v1 , . . . , vk ) f ∈ F for v1 , . . . , vk ∈ E, f ∈ F, and γ ∈

Vk

(E);

(ii) { (ei1 ∧ · · · ∧ eik )fj | i1 < · · · < ik } is a basis of

Vk

(E, F) and thus dim

V

k

 (E, F) =

mn! ; (n − k)!k!

7.2 Determinants, Volumes, and the Hodge Star Operator

393

V (iii) dim( (E, F)) = m2n ; (iv) if B ∈ L(R, F; F), where B(t, f ) = tf and ∧ is the wedge product defined by B, regarded as a map V1 Vk Vk+1 ∧: (E) × (E, F) → (E, F) show that X α∧β = (α ∧ βi ) fi . 1≤i≤m

If E = R3 , F = R2 , α = e1 ∧ e2 − 2e1 ∧ e3 , and


β = e1 ∧ e3 f1 + 2 e2 ∧ e3 f2 − e1 ∧ e2 f3 , compute α ∧ β.  7.1-6. Let {e1 , . . . , ek } and {f1 , . . . , fk } be linearly independent sets of vectors. Show that they span the same k-dimensional subspace iff f1 ∧ · · · ∧ fk = ae1 ∧ · · · ∧ ek , where a 6= 0. (Give a definition of f1 ∧ · · · ∧ fk as part of your answer.) Show that in fact where ϕ : span{e1 , . . . , ek } → span{f1 , ..., fk } Vk is determined by ϕ(ei ) = fi , i = 1, . . . , k. Use this to relate with Gk in Example 3.1.8G. a = det ϕ,

 7.1-7 (P. Chernoff and J. Robbin). Let ∧0 be another wedge product on forms that is associative and differs from the usual wedge product only by a scalar factor; that is, it satisfies α ∧0 β = c(k, l)α ∧ β, where α is a k-form and β is an l-form, c(k, l) is a scalar, and forms of degree zero act as scalars. (i) Prove the “cocycle identity” c(k, l)c(k + l, m) = c(k, l + m)c(l, m). (ii) Define ψ(l) inductively by ψ(0) = ψ(1) = 1 and ψ(l + 1) = c(1, l)ψ(l). Show that the identity c(k, l) = kl ψ(k + l)/ψ(k)ψ(l) holds. Deduce that c(k, l) = c(l, k); that is, ∧0 satisfies α ∧0 β = (−1) β ∧0 α automatically. (iii) Show that c given in (ii) yields an associative wedge product. Aside: the choice ψ(k) = 1/k! converts our wedge product convention to that of Kobayashi and Nomizu [1963].

7.2

Determinants, Volumes, and the Hodge Star Operator

According to linear algebra, the determinant of an n × n matrix is a skew-symmetric function of its rows or columns. Thus, if x1 , . . . , xn ∈ Rn , and we define ω by ω (x1 , . . . , xn ) = det [x1 , . . . , xn ] Vn n (R ). where [x1 , . . . , xn ] denotes the n × n matrix whose columns are x1 , . . . , xn , then ω is an element of We also recall from linear algebra that det[x1 , . . . , xn ] is the oriented volume of the parallelepiped P spanned by x1 , . . . , xn (Figure 7.2.1) and that if xi has components xji , the determinant is given by X det[x1 , . . . , xn ] = (sign σ)x1σ(1) · · · xnσ(s) . σ∈Sn

In this section determinants and volumes are approached from the point of view of exterior algebra. Throughout this section E is assumed to be a finite-dimensional vector space and we denote its dimension by dim E = n.

394

7. Differential Forms

P

Figure 7.2.1. Volume(P ) = det[x1 , x2 , x3 ]

Figure 7.2.2. Image of a cube under a linear map

If ϕ : R3 → R3 is a linear transformation, it is shown in linear algebra that det ϕ is the oriented volume of the image of the unit cube under ϕ (see Figure 7.2.2). In fact det ϕ is a measure of how ϕ changes volumes. In advanced calculus, this fact is the basis for introducing the Jacobian determinant in the change of variables formula for multiple integrals. This background will lead the exposition to the development of the Jacobian determinant of a mapping of manifolds. Definition of the Determinant. element of Tk0 (E) defined by

Recall that the pull-back ϕ∗ α of α ∈ Tk0 (F) by ϕ ∈ L(E, F) is the

(ϕ∗ α) (e1 , . . . , ek ) = α (ϕ (e1 ) , . . . , ϕ (ek )) . If ϕ ∈ GL(E, F), then ϕ∗ = (ϕ−1 )∗ denotes the push-forward. The following proposition is a consequence of the definitions and Proposition 6.1.9. (The same results hold for Banach space valued forms.) Let ϕ ∈ L(E, F) and ψ ∈ L(F, G) Vk Vk (i) ϕ∗ : Tk0 (F) → Tk0 (E) is linear, and ϕ∗ ( (F)) ⊂ (E).

7.2.1 Proposition.

(ii) (ψ ◦ ϕ)∗ = ϕ∗ ◦ ψ ∗ . (iii) If ϕ is the identity, so is ϕ∗ .
(iv) If ϕ ∈ GL(E, F), then ϕ∗ ∈ GL Tk0 (F) , Tk0 (E) , ϕ−1


∗

= ϕ∗

and

−1

(ϕ∗ )

; if ψ ∈ GL (F, G), then (ψ ◦ ϕ)∗ = ψ∗ ◦ ϕ∗ .

= ϕ−1


∗

7.2 Determinants, Volumes, and the Hodge Star Operator

(v) If α ∈

Vk

(F) and β ∈

Vl

395

(F), then ϕ∗ (α ∧ β) = ϕ∗ α ∧ ϕ∗ β.

For example, if β = βa1 ...ak f a1 ∧ · · · ∧ f ak ∈ [Aai ],

and ϕ ∈ L(E, F) is given by the matrix {f1 , . . . , fm } of F, one has ϕ(ei ) = Aai fa , then

Vk

(F)

that is, relative to ordered bases {e1 , . . . ., en } of E and

(ϕ∗ β) = βa1 ···ak ϕ∗ (f a1 ) ∧ · · · ∧ ϕ∗ (f ak ) = =

(sum over a1 < · · · < ak )

(sum over a1 < · · · < ak )

βa1 ···ak Aaj11 ej1 ∧ · · · ∧ Aajkk ejk βa1 ···ak Aaj11 · · · Aajkk ej1 ∧ · · · ∧ ejk k!βa1 ···ak Aaj11 · · · Aajkk ej1 ∧ · · · ∧ ejk , j1

< · · · < jk . V n Recall that ϕ∗ : (E) → (E) is a linear mapping and (E) is one-dimensional. Thus, if ω0 is a ∗ ∗ basis and ω = cω0 , then ϕ ω = cϕ ω0 = bω for some constant b, clearly unique. Vn ∗ 7.2.2 (E) → Vn Definition. Let dim(E) = n and ϕ ∈ L(E, E). The unique constant det ϕ, such that ϕ : (E) satisfies ϕ∗ ω = (det ϕ)ω Vn for all ω ∈ (E) is called the determinant of ϕ. =

Vn

Vn

The definition shows that the determinant does not depend on the choice of basis of E, nor does it depend on a norm on E. To compute det ϕ, choose a basis {e1 , . . . , en } of E with dual basis {e1 , . . . , en }. Let ϕ ∈ L(E, E) have the matrix [Aji ]; that is, ϕ(ei ) = Σ1≤j≤n Aji ej . By Example 7.1.6A, ϕ∗ (e1 ∧ · · · ∧ en )(e1 , . . . , en ) = (e1 ∧ · · · ∧ en )(ϕ(e1 ), . . . , ϕ(en )) = det[ej (ϕ(ei ))] = det[Aji ]. Since (e1 ∧ · · · ∧ en )(e1 , . . . , en ) = 1 we get det ϕ = det[Aji ], the classical expression of the determinant of a matrix with x1 , . . . , xn as columns, where xi has components Aji . Thus the definition of the determinant in Definition 7.2.2 coincides with the classical one. Properties of the Determinant. From properties of pull-back, we deduce corresponding properties of the determinant, all of which are well known from linear algebra. 7.2.3 Proposition.

Let ϕ, ψ ∈ L(E, E). Then

(i) det(ϕ ◦ ψ) = (det ϕ)(det ψ); (ii) if ϕ is the identity, det ϕ = 1; (iii) ϕ is an isomorphism iff det ϕ 6= 0, and in this case det(ϕ−1 ) = (det ϕ)−1 . Proof.

To prove (i), note first that (ϕ ◦ ψ)∗ ω = det(ϕ ◦ ψ)ω; but (ϕ ◦ ψ)∗ ω = (ψ ∗ ◦ ϕ∗ )ω. Hence, ∗

(ϕ ◦ ψ) ω = ψ ∗ (det ϕ)ω = (det ψ)(det ϕ)ω so (i) follows. Part (ii) follows at once from the definition. For (iii), suppose ϕ is an isomorphism with inverse ϕ−1 . Therefore, by (i) and (ii), 1 = det(ϕ ◦ ϕ−1 ) = (det ϕ)(det ϕ−1 ), and, in particular, det ϕ 6= 0. Conversely, if ϕ is not an isomorphism there is an e1 6= 0 satisfying ϕ(e1 ) = 0. Extend to a basis {e1 , e2 , . . . , en }. Then for all n-forms ω, we have (ϕ∗ ω) (e1 , . . . , en ) = ω (0, ϕ (e2 ) , . . . , ϕ (en )) = 0. Hence, det ϕ = 0.



396

7. Differential Forms

In Chapter 2 we saw that if E and F are finite dimensional, one convenient norm giving the topology of L(E, F) is the operator norm:   kϕ(e)k |e 6= 0 kϕk = sup{kϕ(e)k | kek = 1} = sup kek where kek is a norm on E. (See §2.2.) Hence, for any e ∈ E, kϕ(e)k ≤ kϕk kek. 7.2.4 Proposition.

The map det : L(E, E) → R is continuous.

Proof. This is clear from the component formula for the determinant, but let us also give a coordinate free proof. Note that kωk = sup{ |ω(e1 , . . . , en )| | ke1 k = · · · = ken k = 1 } = sup{ |ω(e1 , . . . , en )|/ke1 k · · · ken k | e1 , . . . , en 6= 0 } is a norm on

Vn

(E) and |ω(e1 , . . . , en )| ≤ kωk ke1 k · · · ken k. Then, for ϕ, ψ ∈ L(E, E), | det ϕ − det ψ| kωk = kϕ∗ ω − ψ ∗ ωk = sup{ |ω(ϕ(e1 ), . . . , ϕ(en )) − ω(ψ(e1 ), . . . , ψ(en ))| | ke1 k = · · · = ken k = 1 } ≤ sup{ |ω(ϕ(e1 ) − ψ(e1 ), ϕ(e2 ), . . . , ϕ(en ))| + · · · + |ω(ψ(e1 ), ψ(e2 ), . . . , ϕ(en ) − ψ(en ))| | ke1 k = · · · = ken k = 1 } ≤ kωk kϕ − ψk{kϕkn−1 + kϕkn−2 kψk + · · · + kψkn−1 } ≤ kωk kϕ − ψk(kϕk + kψk)n−1 .

Consequently, | det ϕ − det ψ| ≤ kϕ − ψk(kϕk + kψk)n−1 from which the result follows.



In Chapter 2 we saw that the set of isomorphisms of E to F form an open subset of L(E, F). Using the determinant, we can give an alternate proof in the finite-dimensional case. 7.2.5 Proposition. Suppose that E and F are finite-dimensional and let GL(E, F) denote those ϕ ∈ L(E, F) that are isomorphisms. Then GL(E, F) is an open subset of L(E, F). Proof. If GL(E, F) = ∅, the conclusion is true. Otherwise, there is an isomorphism ψ ∈ GL(E, F). A map ϕ in L(E, F) is an isomorphism if and only if ψ −1 ◦ ϕ is also. This happens precisely when det(ψ −1 ◦ ϕ) 6= 0. Therefore, GL(E, F) is the inverse image of R\{0} under the map taking ϕ to det (ψ −1 ◦ ϕ). Since this is continuous and R\{0} is open, GL(E, F) is also open.  Vn (E) enable us to define orientation or “handedness” of a vector Orientation. The basis elements of space. Vn 7.2.6 Definition. The nonzero elements of the one-dimensional space (E) are called volume elements. If ω1 and ω2 are volume elements, we say ω1 and ω2 are equivalent iff there is a c > 0 such that ω1 = cω2 . An equivalence class [ω] of volume elements on E is called an orientation on E. An oriented vector space (E, [ω]) is a vector space E together with an orientation [ω] on E; [−ω] is called the reverse orientation. A basis {e1 , . . . , en } of the oriented vector space (E, [ω]) is called positively (resp., negatively ) oriented, if ω(e1 , . . . , en ) > 0 (resp., < 0).

7.2 Determinants, Volumes, and the Hodge Star Operator

397

The last statement is independent of the representative of the orientation [ω], for if ω 0 ∈ [ω], then ω 0 = cω for some c > 0, and thus ω 0 (e1 , . . . , en ) and ω(e1 , . . . , en ) have the same sign. Also note that a vector space E has exactly two orientations: one given by selecting an arbitrary dual basis {e1 , . . . , en } and taking [e1 ∧ · · · ∧ en ]; the other is its reverse orientation. This definition of orientation is related to the concept of orientation from calculus as follows. In R3 , a right-handed coordinate system like the one in Figure 7.2.1 is by convention positively oriented, as are all other right-handed systems. On the other hand, any left-handed coordinate system, obtained for example from the one in Figure 7.2.1 by interchanging x1 and x2 , is by convention negatively oriented. Thus one would call a positive orientation in R3 the set of all right-handed coordinate systems. The key to the abstraction of this construction for any vector space lies in the observation that the determinant of the change of ordered basis of two right-handed systems in R3 is always strictly positive. Thus, if E is an n-dimensional vector space, define an equivalence relation on the set of ordered bases in the following way: two bases {e1 , . . . , en } and {e01 , . . . , e0n } are equivalent iff det ϕ > 0, where ϕ ∈ GL(E) is given by ϕ(ei ) = e0i , i = 1, . . . , n. We can relate n-forms to the bases by associating to a basis {e1 , . . . , en } and its dual basis {e1 , . . . , en } the n-form ω = e1 ∧ · · · ∧ en . The following proposition shows that this association gives an identification of the corresponding equivalence classes. 7.2.7 Proposition. ordered bases.

An orientation in a vector space is uniquely determined by an equivalence class of

Proof. If Vn[ω] is an orientation of E there exists a basis {e1 , . . . , en } such that ω(e1 , . . . , en ) 6= 0 since ω 6= 0 in (E). Changing the sign of e1 if necessary, we can find a basis that is positively oriented. Let {e01 , . . . , e0n } be an equivalent basis and ϕ ∈ GL(E), defined by ϕ(ei ) = e0i , i = 1, . . . , n be the change of basis isomorphism. Then if ω 0 ∈ [ω], there exists c > 0 such that ω 0 = cω, so we get ω 0 (e01 , . . . , e0n ) = cω(ϕ(e1 ), . . . , ϕ(en )) = c(ϕ∗ ω)(e1 , . . . , en ) = c(det ϕ)ω(e1 , . . . , en ) > 0. That is, [ω] uniquely determines the equivalence class of {e1 , . . . , en }. Conversely, let {e1 , . . . , en } be a basis of E and let ω = e1 ∧· · ·∧en , where {e1 , . . . , en } is the dual basis. As before, ω 0 (e01 , . . . , e0n ) > 0 for any ω 0 ∈ [ω] and {e01 , . . . , e0n } equivalent to {e1 , . . . , en } ; thus, the equivalence class of the ordered basis {e1 , . . . , en } uniquely determines the orientation [ω].  Volume Elements in Inner Product Spaces. An important point is that to get a particular volume element on E requires additional structure, although the determinant does not. The idea is based on the fact that in R3 the volume of the parallelepiped P = P (x1 , x2 , x3 ) spanned by three positively oriented vectors x1 , x2 , and x3 can be expressed independent of any basis as Vol(P ) = (det[hxi , xj i])1/2 , where [hxi , xj i] denotes the symmetric 3 × 3 matrix whose entries are hxi , xj i. If x1 , x2 , and x3 are negatively oriented, det[hxi , xj i] < 0 and so the formula has to be modified to Vol(P ) = (| det[hxi , xj i]|)1/2 .

(7.2.1)

Densities. The above argument suggests that besides the volumes, there are quantities involving absolute values of volume elements that are also important. This leads to the notion of densities. 7.2.8 Definition. Let α be a real number. A continuous mapping ρ : E × · · · × E → R (n factors of E for E an n-dimensional vector space) is called an α-density if ρ (ϕ (v1 ) , . . . , ϕ (vn )) = | det ϕ|α ρ (v1 , . . . , vn ) , V for all v1 , . . . , vn ∈ E and all ϕ ∈ L(E, E). | |α (E) denote the V Let V α-densities on E. With α = 1, 1-densities on E are simply called densities and | |1 (E) is denoted by | |(E).

398

7. Differential Forms

The determinant of ϕ in this definition is taken with respect to any volume element ofVE. As we saw α in Definition 7.2.2, this is independent V αof the choice of the volume element. Note that | | (E) is onedimensional. Indeed, if ρ1 and ρ2 ∈ | | (E), ρ1 6= 0, and {e1 , . . . , en } is a basis of E, then ρ2 (e1 , . . . , en ) = aρ1 (e1 , . . . , en ), for some constant a ∈ R. If v1 , . . . , vn ∈ E, let vi = ϕ(ei ), defining ϕ ∈ L(E, E). Then ρ2 (v1 , . . . , vn ) = | det ϕ|α ρ2 (e1 , . . . , en ) = a| det ϕ|α ρ1 (e1 , . . . , en ) = aρ1 (v1 , . . . , vn ); that is, ρ2 = aρ1 . Vn V Alpha-densities can be constructed from volume elements as follows. If ω ∈ (E), define |ω|α ∈ | |α (E) by |ω|α (e1 , . . . , en ) = |ω (e1 , . . . , en ) |α Vn V where e1 , . . . , en ∈ E. This association defines a map of (E) to | |α (E). Thus one often uses the notation |ω|α for α-densities. Volume Elements in Inner Product Spaces. We shall construct canonical volume elements (and hence α-densities) for vector spaces carrying a bilinear symmetric nondegenerate covariant two-tensor, and in particular for inner product spaces. First we recall a fact from linear algebra. 7.2.9 Proposition. Let E be an n-dimensional vector space and g = h , i ∈ T20 (E) be symmetric and of rank r; that is, the map e ∈ E 7→ g(e, ·) ∈ E∗ has r-dimensional range. Then there is an ordered basis {e1 , . . . , en } of E with dual basis {e1 , . . . , en } such that g=

r X

ci ei ⊗ ei ,

i=1

where ci = ±1 and r ≤ n, or equivalently, the matrix of g  c1 0 0 · · · 0  0 c2 0 · · · 0   0 0 c3 · · · 0   .. .. .. . . .  . . .. . .   0 0 0 . . . cr   0 0 0 ··· 0   . .. .. . . .  .. . .. . . 0 0 0 ··· 0

is 0 ··· 0 ··· 0 ··· .. .

0 0 0 .. .

0 ··· 0 ··· .. .

0 0 .. .

0 ···

0

       .      

This basis {e1 , . . . , en } is called a g-orthonormal basis. Moreover, the number of basis vectors for which g(ei , ei ) = 1 (resp., g(ei , ei ) = −1) is unique and equals the maximal dimension of any subspace on which g is positive (resp., negative) definite. The number s = the number of +1’s minus the number of −1’s is called the signature of g. The number of −1’s is called the index of g and is denoted Ind(g). Proof (Gram–Schmidt argument).

Since g is symmetric, the following polarization identity holds:

g(e, f ) =

1 g(e + f, e + f ) − g(e − f, e − f ). 4

Thus if g 6= 0, there is an e1 ∈ E such that g(e1 , e1 ) 6= 0. Rescaling, we can assume c1 = g(e1 , e1 ) = ±1. Let E1 be the span of e1 and E2 = { e ∈ E | g(e1 , e) = 0 }. Clearly E1 ∩ E2 = {0}. Also, if z ∈ E, then z − c1 g(z, e1 )e1 ∈ E2 so that E = E1 + E2 and thus E = E1 ⊕ E2 . Now if g 6= 0 on E2 , there is an e2 ∈ E2 such that g(e2 , e2 ) = c2 = ±1. Continue inductively to complete the proof.

7.2 Determinants, Volumes, and the Hodge Star Operator

399

For the second part, in the basis {e1 , . . . , en } just found, let E1 = span { ei | g (ei , ei ) = 1 } ,

E2 = span { ei | g (ei , ei ) = −1 }

and ker g = { e | g (e, e0 ) = 0 for all e0 ∈ E } . Note that ker g = span{ ei | g(ei , ei ) = 0 } and thus E = E1 ⊕ E2 ⊕ ker g. Let F be any subspace of E on which g is positive definite. Then clearly F ∩ ker g = {0}. We also have E2 ∩ F = {0} since any v ∈ E2 ∩ F, v 6= 0, must simultaneously satisfy g(v, v) > 0 and g(v, v) < 0. Thus F ∩ (E2 ⊕ ker g) = {0} and consequently dim F ≤ dim E1 . A similar argument shows that dim E2 is the maximal dimension of any subspace of E on which g is negative definite.  Note that the number of ones in the diagonal representation of g is (r +s)/2 and the number of minus-ones is Ind(g) = (r − s)/2. Nondegeneracy of g means that r = n. In this case e ∈ E may be written e=

g (e, ei ) ei , ci i=1,...,n X

where ci = g(ei , ei ) = ±1 and {ei } is a g-orthonormal basis. For g a positive definite inner product, r = n and Ind(g) = 0; for g a Lorentz inner product r = n and Ind(g) = 1. 7.2.10 Proposition. metric.

Let E be an n-dimensional vector space and g ∈ T20 (E) be nondegenerate and sym-

(i) If [ω] is an orientation of E there exists a unique volume element µ = µ(g) ∈ [ω], called the g-volume, such that µ(e1 , . . . , en ) = 1 for all positively oriented g-orthonormal bases {e1 , . . . , en } of E. In fact, if {e1 , . . . , en } is the dual basis, then µ = e1 ∧ · · · ∧ en . More generally, if {f1 , . . . , fn } is a positively oriented basis with dual basis {f 1 , . . . , f n }, then µ = | det [g (fi , fj )] |1/2 f 1 ∧ · · · ∧ f n . (ii) There is a unique α-density |µ|α , called the g-α-density , with the property that |µ|α (e1 , . . . , en ) = 1 for all g-orthonormal bases {e1 , . . . , en } of E. If {e1 , . . . , en } is the dual basis, then |µ|α = |e1 ∧· · ·∧en |α . More generally, if v1 , . . . , vn ∈ E are positively oriented, then |µ|α (v1 , . . . , vn ) = | det [g (vi , vj )] |α/2 . Proof. First we establish a relation between the determinants of the following three matrices: [g(ei , ej )] = diag(c1 , . . . , cn ) (see Proposition 7.2.9), [g(fi , fj )] for an arbitrary basis {f1 , . . . , fn }, and the matrix representation of ϕ ∈ GL(E) where ϕ(ei ) = fi = Aji ej . By Proposition 7.2.9, we have ! n X
p p g (fi , fj ) = cp e ⊗ e Aki ek , Alj el p=1

= cp δkp δlp Aki Alj = cp Api Apj

(sum on p).

Thus, det[g(fi , fj )] = (det ϕ)2 det[g(ei , ej )]. By Proposition 7.2.9, | det[g(ei , ej )]| = 1.

(7.2.2)

400

7. Differential Forms

(i) Clearly if {e1 , . . . , en } is positively oriented and g-orthonormal, then µ(e1 , . . . , en ) = 1 uniquely determines µ ∈ [ω] by multilinearity. Suppose that {f1 , . . . , fn } is another positively oriented g-orthonormal basis. If ϕ ∈ GL(E) where ϕ(ei ) = fi , i = 1, . . . , n, then by equation (7.2.2) and Proposition 7.2.9, it follows that | det ϕ| = 1 . But 0 < µ(f1 , . . . , fn ) = (ϕ∗ µ)(e1 , . . . , en ) = det ϕ, so that det ϕ = 1. The second statement in (i) follows from the third. For the third statement of (i), note that by equation (7.2.2) µ(f1 , . . . , fn ) = det ϕ = | det[g(fi , fj )]|1/2 . (ii) follows from (i) and the remarks following Definition 7.2.8.



A covariant symmetric nondegenerate two-tensor g on E induces one on the following way. Let α = αi1 ···ik ei1 ∧ · · · ∧ eik (sum over i1 < · · · < ik ) and let

Vk

(E) for every k = 1, . . . , n in

and β = βi1 ···ik ei1 ∧ · · · ∧ eik ∈

Vk

(E),

β i1 ···ik = g i1 j1 g i2 j2 · · · g ik jk βj1 ···jk

(sum over all j1 , . . . , jk ) be the components of the associated contravariant k-tensor, where [g kj ] denotes the inverse of the matrix with entries gij = g(ei , ej ). Then put X 1 X αi1 ···ik β i1 ···ik . (7.2.3) g (k) (α, β) = αi1 ···ik β i1 ···ik = k! i ,··· , i <···
1

k

k

If there is no danger of confusion, we will write hhα, βii = g (k) (α, β). We now show that this definition does not depend on the basis. If {f1 , . . . , fn } is another ordered basis of E, let α = αa0 1 ···ak f a1 ∧ · · · ∧ f ak

and β = βa0 1 ···ak f a1 ∧ · · · ∧ f ak .

The identity map on E has matrix representation relative to the bases {e1 , . . . , en } and {f1 , . . . , fn } given by ei = Aai fa . If B = A−1 we have, by Proposition 6.1.7, αa0 1 ···ak β 0a1 ···ak = αi1 ···ik Bai11 · · · Baikk Aaj11 · · · Aajkk β j1 ···jk = αa1 ···ak δji11 · · · δjikk β i1 ···jk = αi1 ···ik β j1 ···jk . So defined, g (k) is clearly bilinear. It is also symmetric since βi1 ···ik αi1 ···ik = gi1 j1 · · · gik jk β j1 ···jk g i1 l1 . . . g ik lk αl1 ···lk = δjl11 · · · δjlkk αl1 ···lk β j1 ···jk = αj1 ···jk β j1 ···jk , where [g ij ] = [gij ]−1 , and gij = g(ei , ej ). Notice that g (k) is also nondegenerate since if g (k) (α, β) = 0 for all Vk β∈ (E), choosing for β all elements of a basis, show that αi1 ···ik = 0, that is, that α = 0. The following has thus been proved. 7.2.11 Proposition. A nondegenerate symmetric covariant two-tensor g = h , i on the finite-dimensional Vk vector space E induces a similar tensor on (E) for all k = 1, . . . , n. Moreover, if {e1 , . . . , en } is a gorthonormal basis of E in which n X g= ci ei ⊗ ei , ci = ±1, i=1

then the basis { ei1 ∧ · · · ∧ eik | i1 < · · · < ik } is orthonormal with respect to g (k) = h , i, and hei1 ∧ · · · ∧ eik , ei1 ∧ · · · ∧ eik i = ci1 · · · cik (= ±1).

(7.2.4)

7.2 Determinants, Volumes, and the Hodge Star Operator

Hodge Star Operator.

401

This operator will be introduced with the aid of the g-volume µ on E.

7.2.12 Proposition. Let E be an oriented n-dimensional vector space and g = h , i ∈ T20 (E) a given symmetric and nondegenerate tensor. Vk Vn−kLet µ be the corresponding volume element of E. Then there is a unique isomorphism ∗ : (E) → (E) satisfying Vk α ∧ ∗β = hhα, βii µ for α, β ∈ (E). (7.2.5) If {e1 , . . . , en } is a positively oriented g-orthonormal basis of E and {e1 , . . . , en } is its dual basis, then ∗ (eσ(1) ∧ · · · ∧ eσ(k) ) = cσ(1) . . . cσ(k) sign(σ)(eσ(k+1) ∧ · · · ∧ eσ(n) )

(7.2.6)

where σ(1) < · · · < σ(k) and σ(k + 1) < · · · < σ(n). Proof. First uniqueness is proved. Let ∗ satisfy equation (7.2.5) and let β = eσ(1) ∧ · · · ∧ eσ(k) and α be one of the g-orthonormal basis vectors Vk ei1 ∧ · · · ∧ eik of (E), i1 < · · · < ik . By equation (7.2.5), α ∧ ∗β = 0 unless (i1 , . . . , ik ) = (σ(1), . . . , σ(k)). Thus, ∗β = aeσ(k+1) ∧ · · · ∧ eσ(n) for a constant a. But then β ∧ ∗β = a sign(σ)µ and by equation (7.2.4), hβ, βi = cσ(1) . . . cσ(k) . Hence a = cσ(1) . . . cσ(k) sign(σ) and so ∗ must satisfy equation (7.2.6). Thus ∗ is unique. Define ∗ by equation (7.2.6), recalling that eσ(1) ∧· · ·∧eσ(k) for σ(1) < · · · < σ(k) forms a g (k) -orthonormal Vk basis of (E). As before, equation (7.2.5) is then verified using this basis. Clearly ∗ defined by equation Vk Vn−k (7.2.6) is an isomorphism, as it maps the g-orthonormal basis of (E) to that of (E).  7.2.13 Proposition. Let E be an oriented n-dimensional vector space, g = h , i ∈ T20 (E) symmetric and nondegenerate of signature s, V and µ the associated g-volume of E. The Hodge star operator satisfies the k following properties for α, β ∈ (E): α ∧ ∗β = β ∧ ∗α = hhα, βii µ,

(7.2.7)

∗1 = µ, ∗µ = (−1)Ind(g) ,

(7.2.8)

Ind(g)

(−1)

Ind(g)

hh∗α, ∗βii .

∗ ∗ α = (−1) hhα, βii = (−1)

k(n−k)

α,

(7.2.9) (7.2.10)

Proof. Equation (7.2.7) follows from equation (7.2.5) by symmetry of hα, βi. Equations (7.2.8) follow directly from equation (7.2.6), with k = 0, n, respectively, and σ = identity (note that c1 . . . cn = (−1)Ind(g) ). To verify equation (7.2.9), it suffices to take α = eσ(1) ∧ · · · ∧ eσ(k) . By equation (7.2.6), ∗(eσ(k+1) ∧ · · · ∧ eσ(n) ) = beσ(1) ∧ · · · ∧ eσ(k) for a constant b. To find b use equation (7.2.5) with α = β = eσ(k+1) ∧ · · · ∧ eσ(n) to give (see equation (7.2.4)) beσ(k+1) ∧ · · · ∧ eσ(n) ∧ eσ(1) ∧ · · · ∧ eσ(k) = cσ(k+1) . . . cσ(n) µ. Hence b = cσ(k+1) . . . cσ(n) (−1)k(n−k ) sign(σ). Thus, equation (7.2.6) implies ∗ ∗ (eσ(1) ∧ · · · ∧ eσ(n) ) = cσ(1) . . . cσ(k) (sign(σ)) ∗ (eσ(k+1) ∧ · · · ∧ eσ(n) ) = cσ(1) . . . cσ(k) cσ(k+1) . . . cσ(n) (sign(σ))2 (−1)k(n−k) eσ(1) ∧ · · · ∧ eσ(k) = (−1)Ind(g) (−1)k(n−k) eσ(1) ∧ · · · ∧ eσ(k) .

402

7. Differential Forms

Finally for equation (7.2.10), we use equations (7.2.7) and (7.2.9) to give hh∗α, ∗βii µ = ∗α ∧ ∗ ∗ β = (−1)Ind(g) (−1)k(n−k) ∗ α ∧ β = (−1)Ind(g) β ∧ ∗α = (−1)Ind(g) hα, βiµ.



7.2.14 Examples. V1 3 (R ) where R3 has the standard metric and dual basis is given from equation A. The Hodge operator on 1 2 3 2 (7.2.6) by ∗e = e ∧ e , ∗e = −e1 ∧ e3 , and ∗e3 = e1 ∧ e2 . (This is the isomorphism considered in Example 7.1.12B.) B.

Using equation (7.2.5), we compute ∗ in an arbitrary oriented basis. Write ···ik jk+1 ∧ · · · ∧ ejn ∗(ei1 ∧ · · · ∧ eik ) = cji1k+1 ···jn e

(sum over jk+1 < · · · < jn ) and apply equation (7.2.5) with β = ei1 ∧ · · · ∧ eik

and α = ej1 ∧ · · · ∧ ejk

where {j1 , . . . , jk } is a complementary set of indices to {jk+1 , . . . , jn }. One gets   1···n ···ik i1 j1 ik jk 1/2 cij1k+1 = g · · · g | det[g ]| sign . ij ···jn j1 · · · jn Hence ∗ (ei1 ∧ · · · ∧ eik ) = | det[gij ]|1/2

(7.2.11) X

 sign



1···n g i1 j1 · · · g ik jk ejk+1 ∧ · · · ∧ ejn , j1 · · · jn

where the sum is over all (k, n − k) shuffles 

C.

 1···n . j1 · · · jn

In particular, if k = 1, equation (7.2.11) yields ∗ ei = | det([gij ])|1/2

n X

(−1)j−1 g ij e1 ∧ · · · ∧ eˆj ∧ · · · ∧ en

(7.2.12)

j=1

since sign(j1 , j2 , . . . , jn ) = (−1)j−1 , for j2 < · · · < jn , j1 = j, and where eˆj means that ej is deleted. Vk D. From B we can compute the components of ∗α, where α ∈ (E), relative to any oriented basis: write α = αi1 ···ik ei1 ∧ · · · ∧ eik (sum over i1 < · · · < ik ) and apply equation (7.2.11) to give   X 1···n 1/2 (∗α) = | det[gij ]| sign αi1 ···ik g i1 j1 · · · g ik jk ejk+1 ∧ · · · ∧ ejn . j1 · · · jn Hence 1/2

(∗α)jk+1 ···jn = | det[gij ]|

X

αi1 ···ik g

i1 j1

···g

ik jk



1···n sign j1 · · · jn



for jk+1 < · · · < jn and where the sum is over all complementary indices j1 < · · · < jk .

(7.2.13)

7.2 Determinants, Volumes, and the Hodge Star Operator

403

E. Consider R4 with the Lorentz inner product, which in the standard basis {e1 , e2 , e3 , e4 } of R4 has the matrix   1 0 0 0  0 1 0 0     0 0 1 0 . 0 0 0 −1 V1 4 Let {e1 , e2 , e3 , e4 } be the dual basis. The Hodge operator on (R ) is given by

and on

V2

∗e1 = e2 ∧ e3 ∧ e4 ,

∗e2 = −e1 ∧ e3 ∧ e4 ,

∗e3 = e1 ∧ e2 ∧ e4 ,

∗e4 = e1 ∧ e2 ∧ e3 ,

(R4 ) by ∗(e1 ∧ e2 ) = e3 ∧ e4 ,

∗(e1 ∧ e3 ) = −e2 ∧ e4 , ∗(e2 ∧ e3 ) = e1 ∧ e4

∗(e1 ∧ e4 ) = −e2 ∧ e3 , ∗(e2 ∧ e4 ) = e1 ∧ e3 ,

∗(e3 ∧ e4 ) = −e1 ∧ e2 .

If R4 had been endowed with the usual Euclidean inner product, the formulas for ∗e4 , ∗(e1 ∧e4 ), ∗(e2 ∧e4 ), V3 4 and ∗(e3 ∧ e4 ) would have opposite signs. The Hodge ∗ operator on (R ) follows from the formulas on V1 4 V1 4 (R ) and the fact that for β ∈ (R ), ∗∗β = β (from formula (7.2.9)). Thus we obtain

F.

∗(e2 ∧ e3 ∧ e4 ) = e1 ,

∗(e1 ∧ e3 ∧ e4 ) = −e2 ,

∗(e1 ∧ e2 ∧ e4 ) = e3 ,

∗(e1 ∧ e2 ∧ e3 ) = e4 .

If β is a one form and v1 , v2 , . . . , vn is a positively oriented orthonormal basis, then (∗β) (v2 , . . . , vn ) = β (v1 ) .

This follows from equation (7.2.5) taking α = v 1 , the first element in the dual basis and using the orthonormality of v1 , . . . , vn . 

Exercises Let {e1 , e2 , e3 } be the standard dual basis of R3 and V2 3 V1 3 α = e1 ∧ e2 − 2e2 ∧ e3 ∈ (R ), β = 3e1 − e2 + 2e3 ∈ (R ),
and ϕ ∈ L R2 , R3 have the matrix   1 0 0 −1 2 1

 7.2-1.

Compute ϕ∗ α. With the aid of the standard metrics in R2 and R3 , compute ∗α, ∗β, ∗(ϕ∗ α), and ∗(ϕ∗ β). do you get any equalities? Explain.  7.2-2. A map ϕ ∈ L(E, F), where (E, ω), (F, µ) are oriented vector spaces with chosen volume elements, is called volume preserving if ϕ∗ µ = ω. Show that if E and F have the same (finite) dimension, then ϕ is an isomorphism.  7.2-3. A map ϕ ∈ L(E, F), where (E, [µ]) and (F, [ω]) are oriented vector spaces, is called orientation preserving if ϕ∗ µ ∈ [ω]. If dim E = dim F, and ϕ is orientation preserving, show that ϕ is an isomorphism. Given an example for F = E = R3 of an orientation-preserving but not volume-preserving map.

404

7. Differential Forms

 7.2-4. Let E and F be n-dimensional real vector spaces with nondegenerate symmetric two-tensors, g ∈ T20 (E) and h ∈ T20 (F). Then ϕ ∈ L(E, F) is called an isometry if h(ϕ(e), ϕ(e0 )) = g(e, e0 ) for all e, e0 ∈ E. (i) Show that an isometry is an isomorphism. (ii) Consider on E and F the g- and h-volumes µ(g) and µ(h). Show that if ϕ is an orientation-preserving isometry, then ϕ∗ commutes with the Hodge star operator, that is, the following diagram commutes: Vk

∗

(F)

- Vn−k (F)

ϕ∗ Vk ? (E)

-

ϕ∗ V?

n−k

(E)

∗ If ϕ is orientation reversing, show that ∗(ϕ∗ α) = −ϕ∗ (∗α) for α ∈

Vk

(F).

 7.2-5. Let g be an inner product and {f1 , f2 , f3 } be a positively oriented basis of R3 . Denote by the index lowering and raising actions defined by g.

[

and

#

(i) Show that for any vectors u, v ∈ R3  123 [∗(u ∧ v )] = sign | det[g(fa , fb )]|1/2 ui v j g kl fl . ijk [

[

#



(ii) Show that if g is the standard dot-product in R3 the formula in (i) reduces to the cross-product of u and v. (iii) Generalize (i) to define the cross-product of n − 1 vectors u1 , . . . , un−1 in an oriented n-dimensional inner product space (E, g), and find its coordinate expression.  7.2-6. Let E be an n-dimensional oriented vector space and let g ∈ T20 (E) be symmetric and non-degenerate Vk Vn−k of signature s. Using the g-volume, define the Hodge star operator ∗ : (E; F) → (E; F), where F is another finite-dimension vector space by ∗α = (∗αi )fi , Vk where αi ∈ (E), {f1 , . . . , fm } is a basis of F and α = αi fi . Show the following. (i) The definition is independent of the basis of F. Vk (E; F). (ii) ∗∗ = (−1)(n−s)/2+k(n−k) on (iii) If h ∈ T20 (F) and if we let h0 (f, α) = (∗ai )h(f, fi ), then ∗h0 (f, α) = h0 (f, ∗α). V (iv) If ∧ is the wedge product in (E; F) with respect to a given bilinear form on F, then for α, β ∈ Vk (E; F), (∗α) ∧ β = (∗β) ∧ α and α ∧ (∗β) = β ∧ (∗α). Vk (v) Show how g and h induce a symmetric nondegenerate covariant two-tensor on (E; F) and find formulas analogous to equations (7.2.7)–(7.2.10).  7.2-7.

Prove the following identities in R3 using the Hodge star operator: ku × vk2 = kuk2 kvk2 − (u · v)2

and u × (v × w) = (u · w)v − (u · v)w.

7.3 Differential Forms

 7.2-8.

405

(i) Prove the following identity for the Hodge star operator: hh∗α, βii = hhα ∧ β, µii ,

where α ∈

Vk

(E) and β ∈

Vn−k

(E).

(ii) Prove the basic properties of ∗ using (i) as the definition.  7.2-9. Let E be an oriented vector space and S ⊂ T20 (E) be the set of nondegenerate symmetric two–tensors of a fixed signature s. (i) Show that S is open. (ii) Show that the map vol : g 7→ µ(g) assigning to each g ∈ S its g-volume element is differentiable and has derivative at g given by h 7→ (trace h)µ(g)/2.

7.3

Differential Forms

The exterior algebra will now be extended from vector spaces to vector bundles and in particular to the tangent bundle. Exterior Forms on Local Vector Bundles. First of all, we need to consider the action of local bundle maps. As in Chapter 3, U × F denotes a local vector bundle, where U is V open in a Banach space E and F is k a Banach space. From U × F, we construct the local vector bundle U × (F). Now we want to piece these local objects together into a global one. 7.3.1 Definition. Let ϕ :VU × F → U 0 ×V F0 be a local vector bundle map that is an isomorphism on each k k 0 fiber. Then define ϕ∗ : U × (F) → U × (F0 ) by (u, ω) 7→ (ϕ0 (u), ϕu∗ ω), where ϕu is the second factor of ϕ (an isomorphism for each u). 7.3.2 Definition. If ϕ : U × F → U 0 × F0 is a local vector bundle map that is an isomorphism on each fiber, then so is ϕ∗ . Moreover, if ϕ is a local vector bundle isomorphism, so is ϕ∗ . Proof.

This is a special case of Proposition 6.2.4.

The Exterior Algebra of a Vector Bundle. fiberwise. 7.3.3 Definition.



Given a vector bundle, we can form the exterior algebra

Suppose π : E → B is a vector bundle. Define Vk S Vk (E)|A = b∈A (Eb )

where A is a subset of B and Eb = π −1 (b) is the fiber over b ∈ B. Let Vk Vk Vk Vk (π) : (E) → B by (π)(t) = b if t ∈ (Eb ).

Vk

(E)|B =

Vk

(E) and define

7.3.4 Theorem. Assume {E|Ui , ϕi } is a vector bundle atlas for the vector bundle π, where ϕi : E|Ui → Vk Vk Vk Vk Ui0 × F0i . Then { (E)|Ui , ϕi∗ } is a vector bundle atlas of (π) : (E) → B, where ϕi∗ : (E)|Ui → V k 0 0 Ui × (F ) is defined by ϕi∗ |Eb = (ϕi |Eb )∗ . Proof. We must verify VB1 and VB2 in Definition 3.4.4 of a vector bundle. Condition VB1 is clear; for VB2 let ϕi , ϕj be two charts for π, so that ϕi ◦ ϕ−1 is a local vector bundle isomorphism on its domain. j −1 But then, ϕi∗ ◦ ϕ−1 = (ϕ ◦ ϕ ) , which is a local vector bundle isomorphism by Definition 7.3.2.  ∗ i ∗ j j Because ofVthis theorem, the vector bundle structure of π : E → B induces naturally a vector bundle k structure on (E) → B.

406

7. Differential Forms

Differential Forms on Manifolds. We now specialize to the important case when π : E → B is the tangent bundle. If τM : T M → M is the tangent bundle of a manifold M , let Vk Vk Vk Vk (M ) = (T M ) and (τM ), M = Vk Vk so M : (M ) → M is the vector bundle of exterior k forms on the tangent spaces of M . Also, let Vk Ω0 (M ) = F(M ), Ω1 (M ) = T10 (M ), and Ωk (M ) = Γ∞ ( M ), k = 2, 3, . . . . 7.3.5 Proposition. Regarding Tk0 (M ) as an F(M ) module, Ωk (M ) is an F(M ) submodule; that is, Ωk (M ) is a subspace of Tk0 (M ) and if f ∈ F(M ) and α ∈ Ωk (M ), then f α ∈ Ωk (M ). Proof. If α1 , α2 ∈ Ωk (M ) and f ∈ F(M ), then we must show f α1 + α2 ∈ Ωk (M ). This follows from the fact that for each m ∈ M , the exterior algebra on Tm M is a vector space.  V k+l 7.3.6 Proposition. If α ∈ Ωk (M ) and β ∈ Ωl (M ), k, l = 0, 1, . . . , define α ∧ β : M → (M ) by (α ∧ β)(m) = α(m) ∧ β(m). Then α ∧ β ∈ Ωk+l (M ), and ∧ is bilinear and associative. Proof. First, ∧ is bilinear and associative since it is true pointwise. To show α ∧ β is of class C ∞ , consider the local representative of α ∧ β in natural charts. This is a map of the form (α ∧ β)ϕ = B ◦ (αϕ × βϕ ), with αϕ , βϕ ∈ C ∞ and B = ∧, which is bilinear and continuous. Thus (α ∧ β)ϕ is C ∞ by the Leibniz rule.  7.3.7 Definition. Let Ω(M ) denote the direct sum of the spaces Ωk (M ), k = 0, 1, . . . , together with its structure as an (infinite-dimensional ) real vector space and with the multiplication ∧ extended componentwise to Ω(M ). (If dim M = n < ∞, the direct sum need only be taken for k = 0, 1, . . . , n.) We call Ω(M ) the algebra of exterior differential forms on M . Elements of Ωk (M ) are called k-forms. In particular, elements of X∗ (M ) are called one-forms. Note that we generally regard Ω(M ) as a real vector space rather than an F(M ) module (as with T (M )). The reason is that F(M ) = Ω0 (M ) is included in the direct sum, and f ∧ α = f ⊗ α = f α. 7.3.8 Examples. A.

A one-form θ on a manifold M assigns to each m ∈ M a linear functional on Tm M .

B.

A two-form ω on a manifold assigns to each m ∈ M a skew symmetric bilinear map ωm : Tm M × Tm M → R.

C.

For an n-manifold M , a tensor field t ∈ Tsr (M ) has the local expression r t(u) = tij11···i ···js (u)

∂ ∂ ⊗ · · · ⊗ ir ⊗ dxj1 ⊗ · · · ⊗ dxjs , ∂xi1 ∂x

where u ∈ U , (U, ϕ) is a local chart on M , and   ∂ ∂ i1 ir r tij11···i (u) = t dx , · · · , dx , , · · · , (u). ···js ∂xj1 ∂xjs Vk The proof of Proposition 7.1.8 gives the local expression for ω ∈ (M ), namely ω(u) = ωi1 ···ik (u)dxi1 ∧ · · · ∧ dxik , where

 ωi1 ···ik (u) = ω

i1 < · · · < ik ,

∂ ∂ ,··· , i i 1 ∂x ∂x k

 (u).

7.3 Differential Forms

407

D. In Ω(M ), the addition of forms of different degree is “purely formal” as in the case M = E. Thus, for example, if M is a two-manifold (a surface) and (x, y) are local coordinates on U ⊂ M , a typical element of Ω(M ) has the local expression f + g dx + h dy + k dx ∧ dy, for f, g, h, k ∈ F(U ). V1 3 V2 3 E. As in §7.1, we have an isomorphism of vector bundles ∗ : (R ) → (R ) given by dx1 7→ dx2 ∧ dx3 ,

dx2 7→ dx3 ∧ dx1 ,

dx3 7→ dx1 ∧ dx2 .

3 On the other hand, the index lowering action given V1 by3 the standard Riemannian metric on R defines a [ 3 ∗ 3 vector bundle isomorphism : T (R ) → T (R ) = (R ). These two isomorphisms applied pointwise define maps ∗ : X∗ (R3 ) → Ω2 , α 7→ ∗ α

and [

: X(R3 ) → X∗ (R3 ),

X 7→ X [ .

Then Example 7.1.12C implies ∗[(X × Y )[ ] = X [ ∧ Y [ for any vector fields X, Y ∈ X(R3 ) where X × Y denotes the usual cross–product of vector fields on R3 from calculus. That is,
∂
∂ + X 3Y 1 − X 1Y 3 ∂x1 ∂x2
∂ + X 1Y 2 − X 2Y 1 ∂x3

X × Y = X 2Y 3 − X 3Y 2

where X = X i

∂ ∂ and Y = Y i i , i = 1, 2, 3. ∂xi ∂x

F. The wedge product is taken in Ω(M ) in the same way as in the algebraic case. For example, if M = R3 , α = dx1 − x1 dx2 ∈ Ω1 (M ) and β = x2 dx1 ∧ dx3 − dx2 ∧ dx3 , then α ∧ β = (dx1 − x1 dx2 ) ∧ (x2 dx1 ∧ dx3 − dx2 ∧ dx3 ) = 0 − x1 x2 dx2 ∧ dx1 ∧ dx3 − dx1 ∧ dx2 ∧ dx3 + 0 = (x1 x2 − 1)dx1 ∧ dx2 ∧ dx3 .



Pull-back and Push-forward of Forms. We can now extend the pull-back and push-forward operations from the context of vector spaces and linear maps to that of manifolds and nonlinear maps. 7.3.9 Definition. Suppose F : M → N is a C ∞ mapping of manifolds. For ω ∈ Ωk (N ), define F ∗ ω : Vk M→ (M ) by F ∗ ω(m) = (Tm F )∗ ◦ ω ◦ F (m); that is, (F ∗ ω)m (v1 , . . . , vk ) = ωF (m) (Tm F · v1 , . . . , Tm F · vk ), where v1 , . . . , vk ∈ Tm M ; for g ∈ Ω0 (N ), F ∗ g = g ◦ F . We say F ∗ ω is the pull-back of ω by F . (See Figure 7.3.1.) 7.3.10 Proposition.

Let F : M → N and G : N → W be C ∞ mappings of manifolds. Then

(i) F ∗ : Ωk (N ) → Ωk (M ); (ii) (G ◦ F )∗ = F ∗ ◦ G∗ ;

408

7. Differential Forms

F

F(m)

m M

N Figure 7.3.1. Pulling back forms

(iii) if H : M → M is the identity, then H ∗ : Ωk (M ) → Ωk (M ) is the identity; (iv) if F is a diffeomorphism, then F ∗ is an isomorphism and
∗ −1 (F ∗ ) = F −1 ; (v) F ∗ (α ∧ β) = F ∗ α ∧ F ∗ β for α ∈ Ωk (N ) and β ∈ Ωl (N ). Proof. Choose charts (U, ϕ), (V, ψ) of M and N so that F (U ) ⊂ V . Then the local representative Fϕψ = ψ ◦ F ◦ ϕ−1 is of class C ∞ , as is ωψ = (T ψ)∗ ◦ ω ◦ ψ −1 . The local representative of F ∗ ω is (F ∗ ω)ϕ (u) = (T ϕ)∗ ◦ F ∗ ω ◦ ϕ−1 (u) = (Tu Fϕψ )∗ ◦ ωψ ◦ Fϕψ (u) which is of class C ∞ by the composite mapping theorem. For (ii), note that it holds for the local representatives; (iii) follows from the definition; (iv) follows in the usual way from (ii) and (iii); and (v) follows from the corresponding pointwise result.  Vector Bundle Valued Forms (Optional). We close this section with a few optional remarks about vector-bundle-valued forms. As before, the idea is to globalize vector-valued exterior forms. ˆ → B, ρ : Fˆ → B be vector bundles over the same base. Define 7.3.11 Definition. Let π : E V  Vk ˆ ˆ k ˆ (E; F ) = L (E), Fˆ , Vk ˆ the vector bundle with base B of vector bundle homomorphisms over the identity from (E) to Fˆ . If V V k k ˆ = T B, E (T B; Fˆ ) is denoted by (B; Fˆ ) and is called the vector bundle of F -valued k-forms on M . Vk ˆ If F = B × F , we denote it by (B, F ) and call its elements vector-valued k-forms on M . The spaces ˆ Fˆ ), Ωk (B; Fˆ ) and Ωk (B; F ). Finally, Ω(E; ˆ Fˆ ) of sections of these bundles are denoted respectively by Ωk (E; ˆ Fˆ ), k = 1, 2, . . . , n, together with its structure of (resp., Ω(B; Fˆ ), Ω(B, F )) denotes the direct sum of Ωk (E; an infinite-dimensional real vector space and F(B)-module. ˆ Fˆ ) is a smooth assignment to the points b of B of skew symmetric k-linear maps Thus, α ∈ Ωk (E; ˆ ˆ αb : Eb ×· · ·× Eb → Fˆb . In particular, if all manifolds and bundles are finite dimensional, then α ∈ Ωk (M, Rp ) may be uniquely written in the form α = Σi=1,...,p αi ej , where α1 , . . . , αp ∈ Ωk (M ), and {e1 , . . . , ep } is the standard basis of Rp . Thus α ∈ Ωk (E, Rp ) is written in local coordinates as (αi11 ···ik dxi1 ∧ · · · ∧ dxik , . . . , αip1 ···ik dxi1 ∧ · · · ∧ dxik ) for i1 < · · · < ik . Proposition 7.3.10(i)–(iv) and its proof have straightforward generalizations to vectorbundle-valued forms on M . The wedge product requires additional structure to be defined, namely a smooth assignment b 7→ gb of a symmetric bilinear map gb : Fˆb × Fˆb → Fˆb for each b ∈ B. With this structure, Proposition 7.3.10(v) also carries over.

7.3 Differential Forms

409

Exercises  7.3-1. Show that for a vector bundle π : E → B, vector-bundle-valued tensors and forms.  7.3-2.

Vk

(E) is a (smooth) subbundle of Tk0 (E). Generalize to

Let ϕ : R3 → R2 be given by ϕ(x, y, z) = (x2 , yz). For α = v 2 du + dv ∈ Ω1 R2





and β = uvdu ∧ dv ∈ Ω2 R2 ,

compute α ∧ β, ϕ∗ α, ϕ∗ β, and ϕ∗ (α ∧ β).  7.3-3 (E. Cartan’s lemma). Let M be an n-manifold and suppose that α1 , . . . , αk ∈ Ω1 (M ), k ≤ n, are pointwise linearly independent. Show that β 1 , . . . , β k ∈ Ω1 (M ) satisfy Σ1≤i≤k αi ∧ β i = 0 iff there exist C ∞ functions aji ∈ F(M ) satisfying aji = aij such that β j = aji β i . Hint: Work in a local chart and show first that αi can be chosen to be dxi ; the symmetry of the matrix [aji ] follows from antisymmetry of ∧ and the given condition.  7.3-4. A (strong) bundle metric g on a vector bundle π : E → B is a smooth section of L2s (E; R) such that g(b) is an inner product on Eb for every b ∈ B which is (strongly) nondegenerate, that is, eb ∈ Eb 7→ g(b)(eb , ·) ∈ Eb∗ is an isomorphism of Banach spaces. (i) Show that the model of the fiber of E is a Hilbertizable space. (ii) If F ⊂ E is a subbundle of E, show that F ⊥ =

S

b∈B

Fb⊥ is a subbundle of E, where we define

Fb⊥ = { eb ∈ Eb | g(b) (eb , fb ) = 0 for all fb ∈ Fb } . (iii) Show that E = F ⊕ F ⊥ .  7.3-5.

Assume the vector bundle π : E → B has a strong bundle metric.

S (i) If σ : B → E is a smooth nowhere vanishing section of E, let Fb = span{σ(b)}, F = b∈B Fb . Show 1 = R × B. Conclude from that F is a subbundle of E which is isomorphic to the trivial bundle EB 1 1 ⊥ Exercise 7.3-4 that E ⊕ (E ) = E. (ii) Show that a manifold M is parallelizable if and only if T M is isomorphic to a trivial bundle. (iii) Assume that M is a strong Riemannian manifold, admits a nowhere vanishing vector field and that 1 T M ⊕ EM is isomorphic to a trivial bundle. Let N be another manifold of dimension ≥ 1 such that 1 ˆ is trivial. Show that M × N is parallelizable. TN ⊕ E N Hint: Use (i) and pull everything back to M × N by the two projections. (iv) Show that if dim N = 0, the conclusion of (ii) is false. Hint: It is know that the only odd dimensional spheres with trivial tangent bundle are S 1 , S 3 and S 7 . Show that T S 2n−1 has a nowhere vanishing vector field. (v) Show that S a(1) × · · · × S a(n) is parallelizable provided that a(i) ≥ 1, i = 1, . . . , n and at least one a(i) is odd. Hint: Use (iii) and Exercise 3.4-3.

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7. Differential Forms

7.4

The Exterior Derivative, Interior Product, and Lie Derivative

The purpose of this section is to extend the differential of functions to a map d : Ωk (M ) → Ωk+1 (M ) defined for any k. This operator turns out to have marvelous algebraic properties. After studying these we shall show how d is related to the basic operations of div, grad and curl on R3 . Then we develop formulas for the Lie derivative. The Exterior Derivative. We first develop the exterior derivative d for finite-dimensional manifolds. The infinite-dimensional case is discussed in Supplement 7.4A. 7.4.1 Theorem. Let M be an n-dimensional manifold. There is a unique family of mappings dk (U ) : Ωk (U ) → Ωk+1 (U ) (k = 0, 1, 2, . . . , n and U is open in M ) which were merely denote by d, called the exterior derivative on M , such that (i) d is a ∧-antiderivation. That is, d is R linear and for α ∈ Ωk (U ) and β ∈ Ωl (U ), d(α ∧ β) = dα ∧ β + (−1)k α ∧ dβ

(product rule);

(ii) If f ∈ F(U ), then df is as defined in Definition 4.2.5; (iii) d2 = d ◦ d = 0, (i.e., dk+1 (U ) ◦ dk (U ) = 0); (iv) d is natural with respect to restrictions; that is, if U ⊂ V ⊂ M are open and α ∈ Ωk (V ), then d(α|U ) = (dα)|U, that is, or the following diagram commutes: Ωk (V )

|U

d

- Ωk (U ) d ? - Ωk+1 (U )

? Ωk+1 (V ) |U

As usual, condition (iv) means that d is a local operator . Proof.


We first establish uniqueness. Let (U, ϕ) be a chart, where ϕ(u) = x1 , . . . , xn , and let

α = αi1 ···ik dxi1 ∧ · · · ∧ dxik ∈ Ωk (U ), i1 < · · · < ik .
If k = 0, by (ii), the local formula dα = ∂α/∂xi dxi applied to the coordinate functions xi , i = 1, . . . , n shows that the differential of xi is the one-form dxi . From (iii), d(dxi ) = 0, so by (i)
d dxi1 ∧ · · · ∧ dxik = 0. Thus, again by (i), dα =

∂αi1 ···ik i dx ∧ dxi1 ∧ · · · ∧ dxik , ∂xi

(sum over i1 < · · · < ik ),

(7.4.1)

7.4 The Exterior Derivative, Interior Product, & Lie Derivative

411

and so d is uniquely defined on U by properties (i)–(iii), and by (iv) on any open subset of M . For existence, define on every chart (U, ϕ) the operator d by formula (7.4.1). Then (ii) is trivially verified as is R-linearity. If β = βj1 ···jl dxj1 ∧ · · · ∧ dxjl ∈ Ωl (U ), then
d(α ∧ β) = d αi1 ···ik βj1 ···jl dxi1 ∧ · · · ∧ dxik ∧ dxj1 ∧ · · · dxjl   ∂βj1 ···jl ∂αi1 ···ik βj1 ···jl + αi1 ···ik dxi ∧ dxi1 ∧ · · · ∧ dxik = ∂xi ∂xi ∧ dxj1 ∧ · · · ∧ dxjl ∂αi1 ···ik i = dx ∧ dxi1 ∧ · · · ∧ dxik ∧ βj1 ...jl dxj1 ∧ · · · ∧ dxjl ∂xi ∂βj1 ···jl i + (−1)k αi1 ···ik dxi1 ∧ · · · ∧ dxik ∧ dx ∧ dxj1 ∧ · · · ∧ dxjl ∂xi = dα ∧ β + (−1)k α ∧ dβ. and (i) is verified. For (iii), the symmetry of the second partial derivatives shows that d(dα) =

∂ 2 αi1 ···ik i dx ∧ dxj ∧ dxj1 ∧ · · · ∧ dxik = 0, ∂xi ∂xj

i1 < · · · < ik .

Thus, in every chart (U, ϕ), equation (7.4.1) defines the operator d satisfying (i)–(iii). It remains to be shown that these local d’s define an operator d on any open set and (iv) holds. To do this, it is sufficient to show that this definition is chart independent. Let d0 be the operator given by equation (7.4.1) on a chart (U 0 , ϕ0 ), where U 0 ∩ U 6= ∅. Since d0 also satisfies (i)–(iii), and local uniqueness has already been proved, d0 α = dα on U ∩ U 0 . The theorem thus follows.  7.4.2 Corollary.

Let ω ∈ Ωk (U ), where U ⊂ E is open. Then dω(u) (v0 , . . . , vk ) =

k X

i

(−1) Dω(u) · vi (v0 , . . . , vˆi , . . . , vk )

(7.4.2)

i=0

where vˆi denotes that vi is deleted. Also, we denote elements (u, v) of T U merely by v for brevity. (Note that Dω(u) · v ∈ Lka (E, R) since ω : U → Lka (E, R).) Proof. Since we are in the finite dimensional case, we can proceed with a coordinate computation. (An alternative is to check out that d defined by equation (7.4.2) satisfies (i) to (iv).
Checking (i) and (iii) is straightforward but lengthy.) Indeed, if the local coordinates of u are x1 , . . . , xn , ω(u) = ωi1 ···ik (u)dxi1 ∧ · · · ∧ dxik (sum over i1 < · · · < ik ), then Dω(u) · vi =

∂ωi1 ···ik j i1 vi dx ∧ · · · ∧ dxik ∂xj

(where the sum is over all j and i1 < · · · < ik ). From equation (7.4.1), ∂ωi1 . . .ik j dx ∧ dxi1 ∧ · · · ∧ dxik (v0 , . . . , vk ) ∂xj ∂ωi1 . . .ik σ(j) σ(i ) σ(i ) = (sign σ)v0 v1 1 · · · vk k ∂xj

dω (v0 , . . . , vk ) =

(7.4.3)

412

7. Differential Forms

(where the sum is over all i1 < · · · < ik , j, and σ’s satisfying σ(j) < σ(i1 ) < · · · < σ(ik )). The right hand side of equation (7.4.2) is (−1)i

∂ωi1 ···ik j η(i ) η(i ) vi (sign η) v0 1 · · · vˆi η(ij ) · · · vk k ∂xj

(7.4.4)

(where the sum is over all i1 < · · · < ik , j, i, and η’s with η(i1 ) < · · · < η(ik )). Writing σ as a product of a permutation moving j to a designated position and a permutation η, we see that equations (7.4.3) and (7.4.4) coincide.  If M is finite dimensional and α ∈ Ωk (M ) has the local expression α = αi1 ···ik dxi1 ∧ · · · ∧ dxik ∈ Ωk (U ),

i1 < · · · < ik ,

it is straightforward to check that the strict components of dα are given by (dα)j1 ...jk+1 =

k X

(−1)p−1

p=1

∂αj1 ...jp−1 jp+1 ...jk+1 ∂αj1 ...jk + (−1)k ∂xjp ∂xjk+1

for j1 < . . . < jk+1 . Compare with equation (7.4.1) which used nonstrict components. 7.4.3 Examples. A. On R2 , let α = f (x, y)dx + g(x, y)dy. Then dα = df ∧ dx + f d(dx) + dg ∧ dy + gd(dy) by linearity and the product rule. Since d2 = 0, dα = df ∧ dx + dg ∧ dy     ∂f ∂f ∂g ∂g = dx + dy ∧ dx + dx + dy ∧ dy. ∂x ∂y ∂x ∂y Since dx ∧ dx = 0 and dy ∧ dy = 0, this becomes dα =

B.

∂f ∂g dy ∧ dx + dx ∧ dy = ∂y ∂x



∂g ∂f − ∂x ∂y

 dx ∧ dy.

On R3 , let f (x, y, z) be given. Then df =

∂f ∂f ∂f dx + dy + dz, ∂x ∂y ∂z [

so the components of df are those of grad f . That is, (grad f ) = df , where [ is the index lowering operator defined by the standard metric of R3 (see §5.1). C.

On R3 , let F[ = F1 (x, y, z)dx + F2 (x, y, z)dy + F3 (x, y, z)dz. Computing as in Example A yields     ∂F1 ∂F1 ∂F3 ∂F2 [ − dx ∧ dy − − dx ∧ dz dF = ∂x ∂y ∂z ∂x   ∂F3 ∂F2 + − dy ∧ dz. ∂y ∂z

Thus
associated to each vector field G = G1 i + G2 j + G3 k on R3 is the one-form G[ and to this the two-form ∗ G[ by   ∗ G[ = G3 dx ∧ dy − G2 dx ∧ dz + G1 dy ∧ dz, [

where ∗ is the Hodge operator (see §7.2); it is clear the dF[ = ∗(curl F) .

7.4 The Exterior Derivative, Interior Product, & Lie Derivative

D.

413

The divergence is obtained from d by d∗F[ = (div F)dx ∧ dy ∧ dz;

that is,

∗d∗F[ = div F.

Thus associating to a vector field F on R3 the one-form F[ and the two-form d∗F[ , gives rise to the operators curl F and div F. From dF[ = ∗(curl F)[ it is apparent that ddF[ = 0 = d∗(curl F)[ = (div curl F)dx ∧ dy ∧ dz. That is, d2 = 0 gives the well-known vector identity div curl F = 0. Likewise, ddf = 0 becomes d(grad f )[ = 0; that is, ∗(curl grad f )[ = 0. So here d2 = 0 becomes the identity curl grad f = 0.  We summarize the relationship between the operators in vector calculus and differential forms in the table at the end of this section. Mappings and the Exterior Derivative. We will now consider the effect of mappings on the exterior derivative operator d. Recall that Ω(M ) is the direct sum of all the Ωk (M ). 7.4.4 Theorem. Let F : M → N be of class C 1 . Then F ∗ : Ω(N ) → Ω(M ) is a homomorphism of differential algebras; that is, (i) F ∗ (ψ ∧ ω) = F ∗ ψ ∧ F ∗ ω, and (ii) d is natural with respect to mappings; that is, F ∗ (dω) = d (F ∗ ω) , that is, the following diagram commutes: Ωk (M ) 

F∗

d ? Ωk+1 (M ) 

F∗

Ωk (N ) d ? Ωk+1 (N )

Proof. Part (i) was established in Proposition 7.3.10. For (ii), we shall show that if m ∈ M , then there is a neighborhood U of m ∈ M such that d(F ∗ ω|U ) = (F ∗ dω)|U , which is sufficient, as F ∗ and d are both natural with respect to restriction. Let (V, ϕ) be a local chart at F (m) and U a neighborhood of m ∈ M with F (U ) ⊂ V . Then for ω ∈ Ωk (V ), we can write ω = ωi1 ···ik dxi1 ∧ · · · ∧ dxik

(sum over i1 < · · · < ik )

and so dω = ∂i0 ωi1 ···ik dxi0 ∧ dxi1 ∧ · · · ∧ dxik , where ∂i0 = ∂/∂xi0 (sum over i0 and i1 < · · · < ik ) and by (i) F ∗ ω|U = (F ∗ ωi1 ···ik ) F ∗ dxi1 ∧ · · · ∧ F ∗ dxik . If ψ ∈ Ω0 (N ) then d(F ∗ ψ) = F ∗ dψ by the composite mapping theorem, so by (i) and d ◦ d = 0, we get d(F ∗ ω|U ) = F ∗ (dωi1 ···ik ) ∧ F ∗ dxi1 ∧ · · · ∧ F ∗ dxik = F ∗ (dω)|U.



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7. Differential Forms

7.4.5 Corollary. The operator d is natural with respect to push-forward by diffeomorphisms. That is, if F : M → N is a different diffeomorphism, then F∗ dω = dF∗ ω, or the following diagram commutes:

Ωk (M )

F∗

d

- Ωk (N ) d ? - Ωk+1 (N )

? Ωk+1 (M ) F∗

Proof.


∗ Since F∗ = F −1 , the result follows from Theorem 7.4.4(ii).



7.4.6 Corollary. Let X ∈ X(M ). Then d is natural with respect to £X . That is, for ω ∈ Ωk (M ) we have £X ω ∈ Ωk (M ) and d£X ω = £X dω, that is, the following diagram commutes: Ωk (M )

£X - k Ω (M )

d

d ? - Ωk+1 (M )

? Ωk+1 (M ) £X

Proof.

Let Ft be the (local) flow of X. Then we know that d ∗ £X ω (m) = . (Ft ω) (m) dt t=0

Since Ft∗ ω ∈ Ωk (M ), it follows that £X ω ∈ Ωk (M ). Since d commutes with pull-back, we have Ft∗ dω = d (Ft∗ ω). Then, since d is R-linear, it commutes with d/dt and so taking the derivative of this relation at t = 0, we get £X dω = d£X ω.  Interior Products. In Chapter 5, contractions of general tensor fields were studied. For differential forms, contractions play a special role. 7.4.7 Definition.

Let M be a manifold, X ∈ X (M ), and ω ∈ Ωk+1 (M ). Then define iX ω ∈ Tk0 (M ) by iX ω(X1 , . . . , Xk ) = ω(X, X1 , . . . , Xk ).

if ω ∈ Ω0 (M ), we put iX ω = 0. We call iX ω the interior product or contraction of X and ω. (Sometimes X ω is written for iX ω.) 7.4.8 Theorem. f ∈ Ω0 (M ), then

We have iX : Ωk (M ) → Ωk−1 (M ), k = 1, . . . , n, and if α ∈ Ωk (M ), β ∈ Ωl (M ), and

(i) iX is a ∧-antiderivation; that is iX is R-linear and we have the identity iX (α ∧ β) = (iX α) ∧ β + k (−1) α ∧ (iX β); (ii) if X α = f iX α;

7.4 The Exterior Derivative, Interior Product, & Lie Derivative

415

(iii) iX df = £X f ; (iv) £X (α ∧ β) = £X α ∧ β + α ∧ £X β; (v) £X α = iX dα + diX α; (vi) £f X α = f £X α + df ∧ iX α. Proof. That iX α ∈ Ωk−1 (M ) follows from the definitions. For (i), R-linearity is clear. For the second part of (i), write iX (α ∧ β)(X2 , X3 , . . . , Xk+l ) = (α ∧ β)(X, X2 , . . . , Xk+l ) and iX α ∧ β + (−1)k α ∧ iX β =

(k + l − 1)! A (iX α ⊗ β) (k − 1)!l! (k + l − 1)! + (−1)k A(α ⊗ iX β). k!(l − 1)!

Now write out the definition of A in terms of permutations from Definition 7.1.1. The sum over all permutations in the last term can be replaced by the sum over σσ0 , where σ0 is the permutation (2, 3, . . . , k + 1, 1, k + 2, . . . k + l) 7→ (1, 2, 3, . . . , k + l), whose sign is (−1)k . Hence (i) follows. For (ii), we note that α is F(M )-multilinear, and (iii) is just the definition of £X f . Part (iv) follows from the fact that £X is a tensor derivation and commutes with the alternation operator A. (It also follows from the formula for £X in terms of flows.) For (v) we proceed by induction on k. First note that for P k = 0, (iv) reduces to (iii). Now assume that (v) holds for k. Then a (k + 1)-form may be written as dfi ∧ ωi , where ωi is a k form, in some neighborhood of m ∈ M . But £X (df ∧ ω) = £X df ∧ ω + df ∧ £X ω by (iv), so iX d(df ∧ ω) + diX (df ∧ ω) = −iX (df ∧ dω) + d(iX df ∧ ω − df ∧ iX ω) = −iX df ∧ dω + df ∧ iX dω + diX df ∧ ω + iX df ∧ dω + df ∧ diX ω = df ∧ £X ω + d£X f ∧ ω by the inductive assumption and (iii). Since d£X f = £X df , the result follows. Finally, for (vi) we have £f X α = if X dα + dif X α = f iX dα + d (f iX α) = f iX dα + df ∧ iX α + f diX α = f £X α + df ∧ iX α.

(7.4.5) 

Note that proofs of (i), (ii) and (iii) are valid without change on Banach manifolds. Formula (v) £X α = iX dα + diX α,

(7.4.6)

which is sometimes called Cartan’s magic formula, is particularly useful. It can be used in the following way.

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7. Differential Forms

7.4.9 Examples. A. If α is a k-form such that dα = 0 and X is a vector field such that diX α = 0, then Ft∗ α = α, where Ft is the flow of X. Indeed, d ∗ F α = Ft∗ £X α = Ft∗ (iX dα + d(iX α)) = 0. dt t so Ft∗ α is constant in t. Since F0 = identity, Ft∗ α = α for all t. B.

Let M = R3 , suppose div X = 0, and let α = dx ∧ dy ∧ dz. Thus dα = 0. Also, iX α = iX (dx ∧ dy ∧ dz) = X 1 dy ∧ dz − X 2 dx ∧ dz + X 3 dx ∧ dy = ∗X [ .

so diX α = d∗X [ = ∗(div X) = 0. Thus by Example A, Ft∗ (dx ∧ dy ∧ dz) = dx ∧ dy ∧ dz. As we shall see in the next section in a more general context, this means that the flow of X is volume preserving. Of course this can be proved directly as well by differentiating the determinant of the Jacobian matrix of Ft in t (see, for example, Chorin and Marsden [1993]). For related applications to fluid mechanics, see §8.2.  Mappings and the Interior Product. The behavior of contractions under mappings is given by the following proposition. (The statement and proof also hold for Banach manifolds.) 7.4.10 Proposition. Let M and N be manifolds and F : M → N a C 1 mapping. If ω ∈ Ωk (N ), X ∈ X(N ), Y ∈ X(M ), and Y is F -related to X, then iY F ∗ ω = F ∗ iX ω. In particular, if F is a diffeomorphisms, then iF ∗ X F ∗ ω = F ∗ iX ω. That is, interior products are natural with respect to diffeomorphisms and the following diagram commutes: Ωk (N )

F∗

iX ? Ωk−1 (N )

F∗

- Ωk (M ) iF ∗ X ? - Ωk−1 (M )

Similarly for Y ∈ X (M ) we have the following commutative diagram: Ωk (M )

F∗

iY

- Ωk (N ) i F∗ Y ? - Ωk−1 (N )

? Ωk−1 (M ) F∗

7.4 The Exterior Derivative, Interior Product, & Lie Derivative

Proof.

417

Let v1 , . . . , vk−1 ∈ Tm (M ) and n = F (m). Then iY F ∗ ω(m) · (v1 , . . . , vk−1 ) = F ∗ ω(m) · (Y (m), v1 , . . . , vk−1 ) = ω(n) · ((T F ◦ Y )(m), T F (v1 ), . . . , T F (vk−1 )) = ω(n)((X ◦ F )(m), T F (v1 ), . . . , T F (vk−1 )) = iX ω(n) · (T F (v1 ), . . . , T F (vk−1 )) = F ∗ iX ω(m) · (v1 , . . . , vk−1 ).

The Lie Derivative and the Exterior Derivative. Lie derivatives (see Palais [1954]). 7.4.11 Proposition.



The next proposition expresses d in terms of the

Let Xi ∈ X (M ), i = 0, . . . , k, and ω ∈ Ωk (M ). Then we have (£X0 ω)(X1 , . . . , Xk )

(i)

= £X0 (ω(X1 , . . . , Xk )) −

k X

ω(X1 , . . . , £X0 Xi , . . . , Xk )

i=1

and (ii)

dω(X0 , X1 , . . . , Xk ) =

k X

ˆl , . . . , Xk )) (−1)l £Xl (ω(X0 , . . . , X

l=0

X

+

(−1)i+j ω(£Xi (Xj ), X0 , . . . , Xˆi , . . . , Xˆj , . . . , Xk )

0≤i<j≤k

where Xˆi denotes that Xi is deleted. Proof. Part (i) is condition DO1 in Definition 6.3.1. For (ii) we proceed by induction. For k = 0, it is merely dω (X0 ) = £X0 ω. Assume the formula for k − 1. Then if ω ∈ Ωk (M ) we have, by Cartan’s formula (7.4.6) and (i) dω(X0 , X1 , . . . , Xk ) = (iX0 dω)(X1 , . . . , Xk ) = (£X0 ω)(X1 , . . . , Xk ) − (d(iX0 ω))(X1 , · · · , Xk ) = £X0 (ω(X1 , . . . , Xk )) −

k X

ω(X1 , . . . , £X0 Xl , . . . , Xk )

l=1

− (diX0 ω)(X1 , . . . , Xk ) But iX0 ω ∈ Ωk−1 (M ) and we may apply the induction assumption. This gives, after a permutation (d(iX0 ω))(X1 , . . . , Xk ) =

k X

ˆl , . . . , Xj )) (−1)l−1 £Xl (ω(X0 , X1 , . . . , X

l=1

−

X

(−1)i+j ω(£Xi Xj , X0 , X1 , . . . , Xˆi , . . . , Xˆj , . . . , Xk ).

1≤i<j≤k

Substituting this into the foregoing yields the result.



418

7. Differential Forms

Formulas (ii) are quite important in many situations and are worth writing out explicitly for the special cases of one forms and two forms. For γ a one-form and X, Y vector fields, we have dγ(X, Y ) = X[γ(Y )] − Y [γ(X)] − γ([X, Y ]) while for ω a two-form and X, Y, Z vector fields, we have, dω(X, Y, Z) = X[ω(Y, Z)] + Y [ω(Z, X)] + Z[ω(X, Y )] − ω([X, Y ], Z) − ω([Z, X], Y ) − ω([Y, Z], X) Note that the proof of (i) and the first formula in the next corollary holds as well for infinite-dimensional manifolds. 7.4.12 Corollary.

Let X, Y ∈ X(M ). Then [£X , iY ] = i[X,Y ]

and

[£X , £Y ] = £[X,Y ]

In particular, iX ◦ £X = £X ◦ iX . Proof. It is sufficient to check the first formula on any k-form ω ∈ Ωk (U ) and any X1 , . . . , Xk−1 ∈ X(U ) for any open set U of M . Proposition 7.4.11(i) gives (iY £X ω)(X1 , . . . , Xk−1 ) = (£X ω)(Y, X1 , . . . , Xk−1 ) = £X (ω(Y, X1 , . . . , Xk−1 )) −

k−1 X

ω(Y, X1 , . . . [X, Xl ], . . . , Xk−1 ) − ω([X, Y ], X1 , . . . , Xk−1 )

l=1

= £X ((iY ω)(X1 , . . . , Xk−1 )) −

k−1 X


(iY ω)(X1 , . . . , [X, Xl ], . . . , Xk−1 ) − i[X,Y ] ω (X1 , . . . , Xk−1 )

l=1

= (£X iY ω)(X1 , . . . Xk−1 ) − (i[X,Y ] ω)(X1 , . . . , Xk−1 ). One proves [£X , £Y ] = £[X,Y ] using the first relation and Cartan’s formula (7.4.6).



Supplement 7.4A The Exterior derivative on Infinite-dimensional Manifolds Now we discuss the exterior derivative on infinite-dimensional manifolds. Theorem 7.4.1 is rather awkward, primarily because we cannot, without a lot of technicalities, pass from, for example, one-forms to two-forms by linear combinations of decomposable two-forms, that is, two-forms of the type α ∧ β. However, there is a simpler alternative available. 1. Adopt the formula in Proposition 7.4.11(ii) as the definition of d on any open subset of M . Note that at first it is defined as a multilinear function on vector fields and note that £X is already defined. 2. In charts, the equation of Proposition 7.4.11(ii) reduces to the local formula (7.4.2). This or a direct computation shows that d : Ωk (M ) → Ωk+1 (M ) is well defined, depending only on the point values of the vector fields.

7.4 The Exterior Derivative, Interior Product, & Lie Derivative

419

3. One checks the basic properties of d. This can be done in two ways: directly, using the local formula, or using the definition and the following lemma, easily deducible from the Hahn–Banach theorem: if a k-form ω is zero on any set of k vector fields X1 , . . . , Xk ∈ X(U ) for all open sets U in M , then ω = 0. This second method is slightly faster if one first proves formula (7.4.6), which in turn implies Corollary 7.4.12. Proof of formula (7.4.6). Let α be a k-form and X1 , . . . , Xk be a set of k vector fields defined on some open subset of M . Writing X0 = X, we have (iX dα + diX α)(X1 , . . . , Xk ) = dα(X, X1 , . . . , Xk ) + d(iX α)(X1 , . . . , Xk ) =

k X

ˆl , . . . , Xk )) (−1)l £Xl (α(X0 , . . . , X

l=0

+

X

(−1)i+j α(£Xi Xj , X0 , . . . , Xˆi , . . . , Xˆj , . . . , Xk )

0≤i<j≤k

+

k X

ˆl , . . . , Xk )) (−1)l−1 £Xl (α(X0 , X1 , . . . , X

l=1

−

X

(−1)i+j α(£X0 Xj , X0 , X1 , . . . , Xˆi , . . . , Xˆj , . . . , Xk )

1≤i<j≤k

= £X0 (α(X1 , . . . , Xk )) +

k X

(−1)j α(£X0 Xj , X1 , . . . , Xˆj , . . . , Xk )

j=1

= (£X α)(X1 , . . . , Xk )

(by Proposition 7.4.11(ii)).



This and corollary 7.4.12 will allow us to give a proof of the infinite-dimensional version of Corollary 7.4.6 : £X ◦ d = d ◦ £X For functions f this formula is proved as follows. By Proposition 7.4.11(ii), (£X df )(Y ) = £X (df (Y )) − df ([X, Y ]) = X[Y [f ]] − [X, Y ][f ] = Y [X[f ]] = d(X[f ])(Y ) = (d£X f )(Y ). Inductively, assume the formula holds for (k −1)-forms. Then for any k-form α and any vector field Y defined an open subset of M , d£X iY α = £X diY α. Thus by Corollary 7.4.12, iY d£X α = £Y £X α − diY £X α = £X £Y α − £[X,Y ] α + di[X,Y ] α − d£X iY α = £X £Y α − £X diY α − i[X,Y ] dα = £X iY dα − i[X,Y ] dα = iY £X dα. Hence d ◦ £X = £X ◦ d. Next, the remaining properties of d are checked in the following way. R-linearity and Theorem 7.4.1(iv) are immediate consequences of the definition. For Theorem 7.4.1(ii), note that df (X) = iX df = £X f − diX f = £X f = X[f ].

420

7. Differential Forms

To show that d2 = 0, first observe that iX ◦ d ◦ d = £X ◦ d − d ◦ iX ◦ d = d ◦ £X − d ◦ £X + d ◦ d ◦ iX = d ◦ d ◦ iX , so that for any k-form α and any vector fields X1 , . . . , Xk+2 , we have (ddα)(X1 , . . . , Xk+2 ) = iXk+2 . . . iX1 ddα = iXk+2 · · · iX2 ddiX1 α = · · · = iXk+2 ddiXk+1 . . . iX1 α = iXk+2 ddiXk+1 (α(X1 , . . . , Xk )) = 0. The antiderivation property of d is proved by induction using equation (7.4.6) and the antiderivation property for the interior products. Finally, the formula F ∗ ◦ d = d ◦ F ∗ for a map F follows by definition and the properties F ∗ (£X ω) = £X (F ∗ ω), (F ∗ ω)(X1 , . . . , Xk ) = F ∗ (ω(X10 , . . . , Xk0 )), F ∗ [X, Y ] = [X 0 , Y 0 ] if Xi ∼F X 0 , i = 1, . . . , k, X ∼F X 0 , and Y ∼F Y 0 . Thus, with the preceding procedure, d is defined on Banach manifolds and satisfies all the key properties that it does in the finite-dimensional case. These key properties are summarized at the end of this section.

Vector Valued Forms. For vector-valued forms, we adopt, as in the preceding supplement, Palais’ formula from Proposition 7.4.11(ii) as the definition of d on an open subset of M . Note again that this definition uses the fact that £X is defined for vector-valued tensors, and again one has to prove that the local formula in Corollary 7.4.2 holds. Then all properties in the table at the end of this section are verified in the same manner as previously. For vector-valued forms we have an additional formula on Ωk (M ; F ) d◦A=A◦d for any A ∈ L(F, F 0 ). If F is finite dimensional, the definition and properties of d become quite obvious; one notices that if n X ω= ωj fj ∈ Ωk (M ; F ), j=1

where ωj ∈ Ωk (M ) and f1 , . . . , fn is a basis of F , then dω is given by dω =

n X

dωj fj

j=1

and this formula can be taken as the definition of d in this case. This method does not work for vector-bundle valued forms. Additional structure on the bundle is required to be able to lift £X .

7.4 The Exterior Derivative, Interior Product, & Lie Derivative

Closed and Exact Forms and the Poincar´ e Lemma. unification of two well-known facts in vector calculus:

421

The Poincar´e lemma is a generalization and

1. if curl F = 0,then locally F = ∇f ; 2. if div F = 0, then locally F = curl G. 7.4.13 Definition. that ω = dα. 7.4.14 Theorem.

We call ω ∈ Ωk (M ) closed if dω = 0, and exact if there is an α ∈ Ωk−1 (M ) such The following hold:

(i) Every exact form is closed. (ii) Poincar´ e Lemma. If ω is closed, then for each m ∈ M , there is a neighborhood U for which ω|U ∈ Ωk (U ) is exact. Proof. Part (i) is clear since d ◦ d = 0. Using a local chart it is sufficient to consider the case ω ∈ Ωk (U ), where U ⊂ E is a disk about 0 ∈ E, to prove (ii). On U we construct an R-linear mapping H : Ωk (U ) → Ωk−1 (U ) such that d ◦ H + H ◦ d is the identity on Ωk (U ). This will give the result, for dω = 0 implies d(Hω) = ω. For e1 , . . . , ek ∈ E, define Z 1 Hω(u)(e1 , . . . , ek−1 ) = tk−1 ω(tu)(u, e1 , . . . , ek−1 )dt. 0

By Corollary 7.4.2, dHω(u) · (e1 , . . . , ek ) =

k X

(−1)i+1 DHω(u) · ei (e1 , . . . , eˆi , . . . , ek )

i=1

=

k X

tk−1 ω(tu)(ei , e1 , . . . , eˆi , . . . , ek )dt

(−1)

0

i=1

+

1

Z

i+1

k X

(−1)i+1

Z

1

tk Dω(tu) · ei (u, e1 , . . . , eˆi , . . . , ek )dt.

0

i=1

(The interchange of D and the integral is permissible, as ω is smooth and bounded as a function of t ∈ [0, 1].) However, we also have, by Corollary 7.4.2, Z t Hdω(u) · (e1 , . . . , ek ) = tk dω(tu)(u, e1 , . . . , ek )dt 0 Z t = tk Dω(tu) · u(e1 , . . . , ek )dt 0

+

k X i=1

i

1

Z

tk Dω(tu) · ei (u, e1 , . . . , eˆi , . . . , ek )dt.

(−1)

0

Hence 1

Z

ktk−1 ω(tu) · (e1 , . . . , ek )dt

[dHω(u) + HDω(u)](e1 , . . . , ek ) = 0

Z +

1

tk Dω(tu) · u(e1 , . . . , ek )dt

0

Z

1

d k [t ω(tu) · (e1 , . . . , ek )]dt 0 dt = ω(u) · (e1 , . . . , ek ). =



422

7. Differential Forms

There is another proof of the Poncar´e lemma based on the Lie transform method 6.4.7. It will help the reader master the proof of Darboux’ Theorem in §8.1 and is similar in spirit to the proof of Frobenius’ Theorem 4.4.3. Alternative Proof of the Poincar´ e Lemma. Again let U be a ball about 0 in E. Let, for t > 0, Ft (u) = tu. Thus, Ft is a diffeomorphism and, starting at t = 1, is generated by the time-dependent vector field u Xt (u) = ; t that is, F1 (u) = u and dFt (u)/dt = Xt (Ft (u)). Therefore, Cartan’s magic formula, the commuting of pullback and d and the fact that dω = 0, gives d ∗ F ω = Ft∗ £Xt ω = Ft∗ (diX ω + iX dω) = d(Ft∗ iXt ω). dt t For 0 < t0 ≤ 1, we get ω−

Ft∗0 ω

Z

1

=d

Ft∗ iXt ω dt.

t0

Letting t0 → 0, we get ω = dβ, where

1

Z

Ft∗ iXt ω dt.

β= 0

Explicitly, Z βu (v1 , . . . , vk−1 ) =

1

tk−1 ωtu (u, v1 , . . . , vk−1 ) dt.

0

(Note that this formula for β agrees with that in the previous proof.)



Cohomology. It is not true that closed forms are always exact (for example, on R2 \{(0, 0)} or on a sphere—see Exercise 7.4-4). In fact, the quotient groups of closed forms by exact forms (called the deRham cohomology groups of M ) are important algebraic objects attached to a manifold; they are discussed further in §7.6. Below we shall prove that on smoothly contractible manifolds, closed forms are always exact. 7.4.15 Definition. Let t ≥ 1. Two C r maps f, g : M → N are said to be (properly) C r -homotopic, if there exists an  > 0 and a C r (proper ) mapping F : ]−, 1 + [ × M → N such that F (0, m) = f (m), and F (1, m) = g(m) for all m ∈ M . The manifold M is called C r -contractible if there exists a point m0 ∈ M and C r -homotopy F of the constant map m 7→ m0 with the identity map of M ; F is called a C r -contraction of M to m0 The following theorem represents a verification of the homotopy axiom for the deRham cohomology . 7.4.16 Theorem. Let f, g : M → N be two (properly) C r -homotopic maps and α ∈ Ωk (N ) a closed k-form (with compact support) on N . Then g ∗ α − f ∗ α ∈ Ωk (M ) is an exact k-form on M (with compact support). The proof is based on the following. 7.4.17 Lemma (Deformation Lemma).

For a C r -manifold M , let the C r mapping it : M → ]−, 1 + [ × M

be given by it (m) = (t, m). Define H : Ωk+1 (]−, 1 + [ × M ) → Ωk (M )

7.4 The Exterior Derivative, Interior Product, & Lie Derivative

423

by Z Hα =

1

i∗s (i∂/∂t α)ds.

0

Then d ◦ H + H ◦ d = i∗1 − i∗0 . Proof. Since the flow of the vector field ∂/∂t ∈ X(]−, 1 + [ × M ) is given by Fλ (s, m) = (s + λ, m), that is, is+λ = Fλ ◦ is , for any form β ∈ Ωl (]−, 1 + [ × M ) we get d d d d ∗ ∗ ∗ i∗s £∂/∂t β = i∗s Fλ∗ β = i F β = i∗ β = i β. dλ λ=0 dλ λ=0 s λ dλ λ=0 s+λ ds s Therefore, since the integrand in the formula for H is smooth, d and the integral sign commute, so that by Cartan’s formula (7.4.6) and the above formula we get Z

1

i∗s (di∂/∂t + i∂/∂t d)α ds =

d(Hα) + H(dα) = 0

Z = 0

Z

1

i∗s £∂/∂t α ds

0 1

d ∗ i α ds = i∗1 α − i∗0 α. ds s



Proof of Theorem 7.4.16. Define G = H ◦ F ∗ , where H is given in the deformation lemma 7.4.17 and F is the (proper) homotopy between f and g. Since F ∗ commutes with d we get d ◦ G + G ◦ d = g ∗ − f ∗ , so that if the form α ∈ Ωk (N ) is closed (and has compact support), (g ∗ − f ∗ )(α) = d(Gα) (and Gα has compact support).  7.4.18 Lemma (Poincar´e Lemma for Contractible Manifolds). smoothly contractible manifold is exact. Proof.

Any closed k-form (for k > 0) on a

Apply the previous theorem with g = identity on M and f (m) = m0 .



The naturality of the exterior derivative has been investigated by Palais [1959]. He proves the following result. Let M be a connected paracompact n-manifold and assume that A : Ωp (M ) → Ωq (M ) is a linear operator commuting with pull-back, that is, A ◦ ϕ∗ = ϕ∗ ◦ A for any diffeomorphism ϕ : M → M . Then   if 0 ≤ p ≤ n, 0 < q < n, q 6= p, p + 1; 0, A = a(Identity), if 0 < q = p ≤ n;   bd, if 0 ≤ p ≤ n − 1, q = p + 1, for some real constants a, b. If M is compact, then in addition we have   0, if q = 0, 0 < p < n;      c(Identity), if p = q = 0;  A = 0, if q = 0, p = n, M is non-orientable or orientable    and reversible;    d R , if q = 0, p = n, M is orientable and non-reversible, M for some real constants c, d. (Orientability and reversibility will be defined in the next section whereas integration will be the subject of Chapter 7.)

424

7. Differential Forms

Supplement 7.4B Differential Ideals and Pfaffian Systems This box discusses a reformation of the Frobenius theorem in terms of differential ideals in the spirit of E. Cartan. Recall that a subbundle E ⊂ T M is called involutive if for all pairs (X, Y ) of local sections of E defined on some open subset of M , the bracket [X, Y ] is also a local section of E. The subbundle E is called integrable if at every point m ∈ M there is a local submanifold N of M such that Tm N = Em . Frobenius’ theorem states that E is integrable iff it is involutive (see §4.4). Before starting the general theory let us show by a simple example how forms and involutive subbundles are interconnected. Let ω ∈ Ω2 (M ) and assume that Eω = { v ∈ T M | iv ω = 0 } is a subbundle of T M . If X and Y are two sections of Eω then i[X,Y ] ω = £X iY ω − iY £X ω = −iY diX ω − iY iX dω = iX iY dω. This identity shows that if dω = 0, then X, Y ∈ Eω implies that [X, Y ] ∈ Eω and so Eω is integrable. We next refine this argument a little. For any subbundle E, the k-annihilator of E is defined by   α(m)(v1 , . . . , vk ) = 0 Vk 0 E (k) = α ∈ m (M ) . for all v1 , . . . , vk ∈ Em , m ∈ M This is a subbundle of the bundle set U of M and notice that

Vk

(M ) of k-forms. Denote by Γ(U, E) the C ∞ sections of E over the open V I(E) = ⊕0≤k<∞ (M, E 0 (k))

is an ideal of ω(M ); that is, if ω1 , ω2 , ∈ I(E) and ρ ∈ Ω(M ), then ω1 + ω2 ∈ I(E) and ρ ∧ ω1 ∈ I(E). 7.4.19 Proposition. Let E be a subbundle of T M . Then E is involutive if for all open subsets U of M and all ω ∈ Γ(U, E 0 (1)), we have dω ∈ Γ(U, E 0 (2)). If E is involutive, then ω ∈ Γ(U, E 0 (k)) implies dω ∈ Γ(U, E 0 (k + 1)). Proof.

For any α ∈ Γ(U, E 0 (1)) and X, Y ∈ Γ(U, E), Proposition 7.4.11(ii) shows that dα(X, Y ) = X[α(Y )] − Y [α(X)] − α([X, Y ]) = −α([X, Y ]).

Thus, E is involutive iff dα(X, Y ) = 0; that is, dα ∈ Γ(U, E 0 (2)).



The Frobenius theorem in terms of differential forms takes the following form. 7.4.20 Corollary. A subbundle E ⊂ T M is integrable if, for all open subsets U of M , ω ∈ Γ(U, E 0 (1)) implies dω ∈ Γ(U, E 0 (2)). The following considerations are strictly finite dimensional. They can be generalized to infinite-dimensional manifolds under suitable splitting assumptions. We restrict ourselves to the finite-dimensional situation due to their importance in applications and for simplicity of presentation. 7.4.21 Definition. Let M be an n-manifold and I ⊂ Ω(M ) be an ideal. We say that I is locally generated by n − k independent one-forms, if every point of M has a neighborhood U and n − k pointwise linearly independent one-forms ω1 , . . . , ωn−k ∈ Ω1 (U ) such that: Pn−k (i) if ω ∈ I, then ω|U = i=1 Θi ∧ ωi for some Θi ∈ Ω(M ); Pn−k (ii) if ω ∈ Ω(M ) and M is covered by open sets U such that for each U in this cover, ω|U = i=1 Θi ∧ ωi for some Θi ∈ Ω(M ), then ω ∈ I.

7.4 The Exterior Derivative, Interior Product, & Lie Derivative

425

The ideal I ⊂ ω(M ) is called a differential ideal if dI ⊂ I. Finitely generated ideals of Ω(M ) are characterized by being of the form I(E). More precisely, we have the following. 7.4.22 Proposition. Let I be an ideal of ω(M ) and let n = dim(M ). the ideal I is locally generated by n − k linearly independent one-forms iff there exists a subbundle E ⊂ T M with k-dimensional fiber such that I = I(E). Moreover, the bundle E is uniquely determined by I. Proof. If E has k-dimensional fiber, let Xn−l+1 , . . . , Xn be a local basis of Γ(U, E). Complete this to a basis of X(U ) and let ωi ∈ Ω1 (U ) be the dual basis. Then clearly ω1 , . . . , xn−k are linearly independent and locally generate I(E). Conversely, let ω1 , . . . , ωn−k generate I over U and define Em = { v ∈ Tm M | ωi (m)(v) = 0, 1, . . . , n − k }. Em is clearly independent of the generators of I over U so that E = ∩m∈M Em is a subbundle of T M . It is straightforward to check that I = I(E). Finally, E is unique since E 6= E 0 implies I(E) 6= I(E 0 ) by construction.  Different ideals are characterized among finitely generated ones by the following. 7.4.23 Proposition. Let I be an ideal of Ω(M ) locally generated by n − k linearly independent forms ω1 , . . . , ωw−k ∈ Ω1 (U ), n = dim(M ), and let ω1 ∧ · · · ∧ ωw−k = ω ∈ Ωn−k (U ). Then the following are equivalent: (i) I is a differential ideal ; Pn−k (ii) dω = j=1 ω ij ∧ ωj for some ωij ∈ Ωl (U ) and for every U, as in the hypothesis; (iii) dωi ∧ ω = 0 for all open sets U, as in the hypothesis; (iv) there exists Θ ∈ Ω1 (U ) such that dω = Θ ∧ ω for all open sets U, as in the hypothesis. Proof. That conditions (i) and (ii) are equivalent and (ii) implies (iv) follows from the definitions. Condition (iv) means that n−k X

(−1)i dωi ∧ ω1 ∧ · · · ∧ ω ˆ i ∧ · · · ∧ ωn−k = Θ ∧ ω1 ∧ · · · ∧ ωn−k ,

i=1

so that multiplying by ωi we get (iii). Finally, we show that (iii) implies (ii). Let ω1 , . . . , ωn ∈ Ω1 (U ) be a basis that ω1 , . . . , ωn−k generates for I over U . Then X dωi = αijl ωj ∧ ωl , where αijl ∈ F(U ). j
But X

0 = dωi ∧ ω =

αijl ωj ∧ ωl ∧ ω1 ∧ · · · ∧ ωn−k

n−k<j
and thus αijl = 0 for n − k < j < l. Hence dωi =

n−k X j=1

 −

n X l=j+1

 αijl ωl  ∧ ωj .



426

7. Differential Forms

Assembling the preceding results, we get the following version of the Frobenius theorem. 7.4.24 Theorem. Let M be an n-manifold and E ⊂ T M be a subbundle with k-dimensional fiber, and I(E) the associated ideal. The following are equivalent: (i) E is integrable; (ii) E is involutive; (iii) I(E) is a differential ideal locally generated by n − k linearly independent one-forms ω1 , . . . , ωn−k ∈ Ω1 (U ); (iv) for every point of M there exists an open set U and ω1 , . . . , ωn−k ∈ Ω1 (U ) generating I(E) such that dωi =

n−k X

ωij ∧ ωj

for some ωij ∈ Ω1 (U );

j=1

(v) same as (iv) but where the ωi satisfy: dωi ∧ ω1 ∧ · · · ∧ ωn−k = 0; (vi) same as (iv) but with the condition on ωi being: there exists Θ ∈ Ω1 (U ) such that dω = Θ ∧ ω, where ω = ω1 ∧ · · · ∧ ωn−k . 7.4.25 Examples. A.

In classical texts (such as Cartan [1945] and Flanders [1963]), a system of equations ω1 = 0, . . . , ωn−k = 0

where ωi ∈ Ω1 (U ) and U ⊂ Rn

is called Pfaffian system. A solution to this system is a k-dimensional submanifold N of U given by xi = xi (u1 , . . . , uk ) such that if one substitutes these values of xi in the system, the result is identically zero. Geometrically, this means that ω1 , . . . , ωn−k annihilate T N . Thus, finding solutions of the Pfaffian system reverts to finding integral manifolds of the subbundle E = { v ∈ T U | ω(v) = 0, i = 1, . . . , n − k } for which Frobenius’ theorem is applicable; thus we must have dωi ∧ ω1 ∧ · · · ∧ ωn−k = 0. This condition is equivalent to the existence of smooth functions aij , bj on U such that ωi =

n−k X

aij dbj .

j=1

To see this, recall that by the Frobenius theorem there are local coordinates b1 , . . . , bn on U such that the integral of E are given by b1 = constant, so that dbi , i = 1, . . . , n − k annihilate P the tangent spaces to these submanifolds. Thus the ideal I generated by db1 , . . . , dbn−k ; that is, ωi = j=1,...,n−k aij dbj for some smooth functions aij on U .

7.4 The Exterior Derivative, Interior Product, & Lie Derivative

B.

427

Let us analyze the case of one Pfaffian equation in R2 . Let ω = P (x, y)dx + Q(x, y)dy ∈ Ω1 (R2 )

using standard (x, y) coordinates. We seek a solution to ω = 0. This is equivalent to dy/dx = −P (x, y)/Q(x, y), so existence and uniqueness of solutions for ordinary differential equations assures the local existence of a function f (x, y) such that f (x, y) = constant give the integral curves y(x). In other words, f (x, y) = constant is an integral manifold of ω = 0. The same result could have been obtained by means of the Frobenius theorem. Since dω ∧ ω ∈ Ω3 (R2 ), we get dω ∧ ω = 0, so integral manifolds exist and are unique. In texts on differential equations, this problem is often solved with the aid of integrating factors. More precisely, if ω is not (locally) exact, can a function f and function g, called an integrating factor , be found, such that gω = df ? The answer is “yes” by Theorem 7.4.24(iii) for choosing f as above, E = ker df locally. Thus, g is found by solving the partial differential equation, ∂(gQ) ∂(gP ) = . ∂y ∂x This always has a solution and the connection between g and f is given by g=

1 ∂f 1 ∂f = , P ∂x Q ∂y

f (x, y) = constant being the solution of ω = 0. C. Let us analyze a Pfaffian equation ω = 0 in Rn . As in Example B, we would like to be able to write gω = df with df 6= 0 on U ⊂ Rr , for then f (x1 , . . . , xn ) = constant gives the (n − 1)-dimensional integral manifolds; that is, the bundle defined by ω integrable. Conversely, if the bundle defined by ω is integrable, then by Example B, gω = df . Integrability is (by the Frobenius theorem) equivalent to dω ∧ ω = 0, which, as we have seen in Example B, is always verified for n = 2. For n ≥ 3, however, this is a genuine condition. If n = 3, let ω = P (x, y, z)dx + Q(x, y, z)dy + R(x, y, z)dz. Then  ∂R ∂Q dω ∧ ω = (P − ) ∂y ∂z  ∂R ∂Q ∂P ∂P − ) + R( − ) dx ∧ dy ∧ dz; + Q( ∂z ∂x ∂x ∂y thus, ω = 0 is integrable iff the term in the square brackets vanishes. D. The Frobenius’ theorem is often used in overdetermined systems of partial differential equations to answer the question of existence and uniqueness of solutions. Consider for instance the following system of Mayer [1872] in Rp+q = {(x1 , . . . , xp , y 1 , . . . , y q )}: dy α 1 p 1 q = Aα i (x , . . . , x , y , . . . , y ), dxi

i = 1, . . . , p, α = 1, . . . , q.

We ask whether there is a solution y = f (x, c) for any choice of initial conditions c such that f (0, c) = c. The system is equivalent to the following Pfaffian system: i ω α = dy α − Aα i dx = 0.

Since the existence of a solution is equivalent to the existence of p-dimensional integral manifolds, Frobenius’ theorem asserts that the existence and uniqueness is equivalent to X dω α = ω αβ ∧ ω β β=1,...,q

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7. Differential Forms

for some one-forms ω αβ . A straightforward computation shows that α dω α = Cjk dxj ∧ dxk +

where α Cjk =

∂Aα i dxi ∧ ω β , ∂y β

∂Aα ∂Aα ∂Aα β ∂Aα j k k β − + A − A . ∂xk ∂xj ∂y β k ∂y β j

Since dx1 , . . . , dxp , ω 1 , . . . , ω q are a basis of Ω1 (Rp+q ), we see that X α dω α = ω αβ ∧ ω b for some one-forms ω αβ iff Cjk = 0. β=1,...,q α Thus the Mayer system is integrable iff Cjk = 0.



In §8.4 and §8.5 we shall give some applications of Frobenius’ theorem to problems in constraints and control theory. Many of these applications may alternatively be understood in terms of Pfaffian systems; see, for example, Hermann [1977] (Chapter 18).

VECTOR CALCULUS AND DIFFERENTIAL FORMS 1. Sharp and Flat (Using standard coordinates in R3 ) (a) v [ = v 1 dx + v 2 dy + v 3 dz = one-form corresponding to the vector v = v 1 e1 + v 2 e2 + v 3 e3 (b) α# = α1 e1 + α2 e2 + α3 e3 = vector corresponding to the one-form α = α1 dx + α2 dy + α3 dz 2. Hodge Star Operator (equation (7.2.6)), (a) ∗1 = dx ∧ dy ∧ dz (b) ∗dx = dy ∧ dz, ∗dy = dx ∧ dz ∗ dz = dx ∧ dy (c) ∗(dy ∧ dz) = dx, ∗(dx ∧ dz) = −dy, ∗(dx ∧ dy) = dz (d) ∗(dx ∧ dy ∧ dz) = 1 3. Cross Product and Dot Product
#  (a) v × w = ∗ v[ ∧ w[ (b) (v · w)dx ∧ dy ∧ dz = v[ ∧ ∗w[ ; that is, v · w = ∗ v [ ∧ ∗w[




4. Gradient #

∇f = grad f = (df ) 5. Divergence

  ∇ · F = div F = ∗d ∗F[ 6. Curl

h  i# ∇ × F = curl F = ∗ dF[

7.4 The Exterior Derivative, Interior Product, & Lie Derivative

429

IDENTITIES FOR VECTOR FIELDS AND FORMS 1. Vector fields on M with the bracket [X, Y ] form a Lie algebra; that is, [X, Y ] is real bilinear, skew symmetric, and Jacobi’s identity holds: [[X, Y ], Z] + [[Z, X], Y ] + [[Y, Z], X] = 0. Locally, [X, Y ] = DY · X − DX · Y and on functions, [X, Y ][f ] = X[Y [f ]] − Y [X[f ]]. 2. For diffeomorphisms ϕ, ψ, we have ϕ∗ [X, Y ] = [ϕ∗ X, ψ∗ Y ]

and

(ϕ ◦ ψ)∗ X = ϕ∗ ψ∗ X.

3. The forms on a manifold are a real associative algebra with ∧ as multiplication. Furthermore, α ∧ β = (−1)kl β ∧ α for k- and l-forms α and β, respectively. 4. For maps, ϕ, ψ, we have ϕ∗ (α ∧ β) = ϕ∗ α ∧ ϕ∗ β,

(ϕ ◦ ψ)∗ α = ψ ∗ ϕ∗ α.

5. d is a real linear map on forms and ddα = 0,

d(α ∧ β) = dα ∧ β + (−1)k α ∧ dβ

for α a k-form.

6. For α a k-form and X0 , . . . , Xk vector fields: dα(X0 , . . . , Xk ) k X

=

(−1)i Xi [α(X0 , . . . , Xˆi , . . . , Xk )]

i=0

X

+

(−1)i+j α([Xi , Xj ], X0 , . . . , Xˆi , . . . , Xˆj , . . . , Xk ).

0≤i<j≤k

If M is finite dimensional and α =

P

(dω)j1 ...jk+1 =

αi1 ...ik dxi1 ∧ · · · ∧ . . . dxik , (with a sum over i1 < · · · < ik ), then

k X

(−1)p−1

p=1

+ (−1)k

∂ αj ...j j ...j ∂xjp 1 p−1 p+1 k+1

∂ αj ...j , ∂xjk+1 1 k

for j1 < · · · < jk+1 .

Locally, dω(x)(v0 , . . . , vk ) =

k X i=0

∗

∗

7. For a map ϕ, ϕ dα = dϕ α.

(−1)i Dω(x) · vi (v0 , . . . , vˆi , . . . , vk )

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7. Differential Forms

e lemma.) If dα = 0, then α is locally exact; that is, there is a neighborhood U about each 8. (Poincar´ point on which α = dβ for some form β defined on U . The same result holds globally on a contractible manifold. 9. iX α is real bilinear in X, α and for h : M → R, ihX α = hiX α = iX hα. Also iX iX α = 0, and iX (α ∧ β) = iX α ∧ β + (−1)k α ∧ iX β for α a k-form. 10. For a diffeomorphism ϕ, we have ϕ∗ iX α = iϕ∗ X ϕ∗ α; if f : M → N is a mapping and Y is f -related to X, then iY f ∗ α = f ∗ iX α. 11. £X α is real bilinear in X, α and £X (α ∧ β) = £X α ∧ β + α ∧ £X β. 12. (Cartan’s Magic Formula.) £X α = diX α + iX dα. 13. For a diffeomorphism ϕ, ϕ∗ £X α = £ϕ∗ X ϕ∗ α; if f : M → N is a mapping and Y is f -related to X, then £Y f ∗ α = f ∗ £X α. 14.

(£X α)(X1 , . . . , Xk ) = X[α(X1 , . . . , Xk )] −

k X

α(X1 , . . . , [X, Xi ], . . . , Xk ).

i=1

Locally, (£X α)x · (v1 , . . . , vk ) = DαX · X(x) · (v1 , . . . , vk ) n X + αx (v1 , . . . , DXx · vi , . . . , vk ). k=1

15. The following identities hold: (a) £f X α = f £X α + df ∧ iX α. (b) £[X,Y ] α = £X £Y α − £Y £X α. (c) i[X,Y ] α = £X iY α − iY £X α. (d) £X dα = d£X α. (e) £X iX α = iX £X α.

7.4 The Exterior Derivative, Interior Product, & Lie Derivative

431

16. If M is a finite dimensional manifold, X = X l ∂/∂xl and α = αi1 ...ik dxi1 ∧ · · · ∧ dxik ,

i1 < · · · < ik ,

the following local formulas hold: ∂αi1 ...ik l dx ∧ dxi1 ∧ · · · ∧ dxik , ∂xl iX α = X l αli2 ...ik dxi2 ∧ · · · ∧ dxik , ∂αi1 ...ik i1 £X α = X l dx ∧ · · · ∧ dxik ∂xl ∂X i1 l +αi1 i2 ...ik dx ∧ dxi2 ∧ · · · ∧ dxik ∂xl ∂X i2 i1 +αi2 i2 ...ik dx ∧ dxl ∧ dxi3 ∧ · · · ∧ dxik + . . . ∂xl dα =

The Maurer–Cartan Equations We close this section with a proof of the Maurer–Cartan structure equations on a Lie group G. Define λ, ρ ∈ Ω1 (G; g), the space of g-valued one-forms on G, by λ(ug ) = Tg Lg−1 (ug ),

ρ(ug ) = Tg Rg−1 (ug ).

Thus, λ and ρ are Lie-algebra-valued one-forms on G that are defined by left and right translation to the identity, respectively. Define the two-form [λ, λ] by [λ, λ](u, v) = [λ(u), λ(v)], and similarly for [ρ, ρ]. 7.4.26 Theorem (Maurer–Cartan Structure Equations). dλ + [λ, λ] = 0,

dρ − [ρ, ρ] = 0.

Proof. We use identity 6 from the preceding table. Let X, Y ∈ X(G) and let ξ = Tg Lg−1 (X(g)) and η = Tg Lg−1 (Y (g)) for fixed g ∈ G. Recalling that Xξ denotes the left invaraint vector field on G equalling ξ at the identity, we have (dλ)(Xξ , Xη ) = Xξ [λ(Xη )] − Xη [λ(Xξ )] − λ([Xξ , Xη ]). Since λ(Xη )(h) = Th Lh−1 (Xη (h)) = η is constant, the first term vanishes. Similarly, the second term vanishes. The third term equals λ([Xξ , Xη ]) = λ(X[ξ,η] ) = [ξ, η], and hence (dλ)(Xξ , Xη ) = −[ξ, η]. Therefore, (dλ + [λ, λ]) (Xξ , Xη ) = −[ξ, η] + [λ, λ](Xξ , Xη ) = −[ξ, η] + [λ(Xξ ), λ(Xη )] = −[ξ, η] + [ξ, η] = 0.

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7. Differential Forms

This proves that (dλ + [λ, λ]) (X, Y )(g) = 0. Since g ∈ G was arbitrary as well as X and Y , it follows that dλ + [λ, λ] = 0. The second relation is proved in the same way but working with the right-invariant vector fields Yξ , Yη . The sign in front of the second term changes, since [Yξ , Yη ] = Y−[ξ,η] .  Remark. If α is a (0, k)-tensor with values in a Banach space E1 , and β is a (0, l)-tensor with values in a Banach space E2 , and if B : E1 × E2 → E3 is a bilinear map, then replacing multiplication of reals in the formula for the tensor and wedge product by the multiplication B, we get an E3 -valued (0, k + l)-tensor on M . In this way, if α ∈ Ωk (M, E1 ) and β ∈ Ωl (M, E2 ), then



 (k + l)! A(α ⊗ β) ∈ Ωk+l (M, E3 ). k!l!

We call this expression the wedge product associated to B and denote it either by α ∧B β or B ∧ (α, β). In particular, if E1 = E2 = E3 = g and B = [ , ] is the Lie algebra bracket, then for α, β ∈ Ω1 (M ; g), we have [α, β]∧ (u, v) = [α(u), β(v)] − [α(v), β(u)] = −[β, α]∧ (u, v) for any vectors u, v tangent to M . Thus, alternatively, one can write the structure equations as dλ + 21 [λ, λ]∧ = 0,

dρ − 21 [ρ, ρ]∧ = 0.



Exercises  7.4-1.

Compute the exterior derivative of the following differential forms on R3 : α = x3 dx + y 3 dx ∧ dy + xyzdx ∧ dz; β = 3dx dx ∧ dy + 9 cos(xy)dx ∧ dy ∧ dz.

 7.4-2. Using Examples 7.4.3 and the properties of d and ∗, prove the following formulas in R3 for f, g : R3 → R and F, G ∈ X(R3 ): (i) grad(f g) = f (grad g) + g(grad f ). (ii) curl (f F) = (grad f ) × F + f (curl F). (iii) div (f F) = grad(f ) · F + f div F. (iv) div (F × G) = G · curl F − F · curl G. (v) £F G = (F · ∇)G − (G · ∇)F = F div G − G div F − curl(F × G). (vi) curl(F × G) = (div G)F − (div F)G + (G · ∇)F − (F · ∇)G. (vii) curl(curl F) = grad(div F) − ∆F, where (∆F)i = is the usual Laplacian.

∂2F i ∂2F i ∂2F i + + ∂x2 ∂y 2 ∂z 2

7.4 The Exterior Derivative, Interior Product, & Lie Derivative

433

(viii) ∇(F · F) = 2(F · ∇)F + 2F × curl F.  7.4-3. Let ϕ : S 1 × R+ → R2 be the polar coordinate mapping defined by ϕ(θ, r) = (r cos θ, r sin θ). Compute ϕ∗ (dx ∧ dy) from the definitions and verify that it equals d(ϕ∗ x) ∧ d(ϕ∗ y).  7.4-4. On S 1 find a closed one-form α that is not exact. Hint: On R2 \{0} consider α = (ydx − xdy)/(x2 + y 2 ).  7.4-5.

Show that the following properties uniquely characterize iX on finite-dimensional manifolds

(i) iX : Ωk (M ) → Ωk−1 (M ) is a ∧ antiderivation; (ii) iX f = 0 for f ∈ F(M ); (iii) iX ω = ω(X) for ω ∈ Ω1 (M ); (iv) iX is natural with respect to restrictions. Use these properties to show that i[X,Y ] = £X iY − iY £X . Finally, show iX ◦ iX = 0.  7.4-6. Show that a derivation mapping Ωk (M ) to Ωk+1 (M ) for all k ≥ 0 is zero (note that d and iX are anti derivations).  7.4-7.

Let s : T 2 M → T 2 M be the canonical involution of the second tangent bundle (see Exercise 3.4-4).

(i) If X is a vector field on M , show that s ◦ T X is a vector field on T M . (ii) If Ft is the flow on X, prove that T Ft is a flow on T M generated by s ◦ T X. (iii) If µ is a one-form on M , µ0 : T M → R is the corresponding function, and w ∈ T 2 M , then show that dµ0 (sw) = dµ(τT M (w), T τM (w)) + dµ0 (w).  7.4-8. Prove the following relative Poincar´ e lemma. Let ω be a closed k-form on a manifold M and let N ⊂ M be a closed submanifold. Assume that the pull-back of ω to N is zero. Then there is a (k − 1)-form α on a neighborhood N such that dα = ω and α vanishes on N . If ω vanishes on N , then α can be chosen so that all its first partial derivatives vanish on M . Hint: Let ϕt be a homotopy of a neighborhood of N to N and construct an H operator as in the Poncar´e lemma using ϕt .  7.4-9 (Angular variables). Let S 1 denote the circle identified as S 1 = R/(2π) = { x ∈ C | |z| = 1 }. Let γ : R → S 1 ; x 7→ eix be the exponential map. Show that γ induces a isomorphism T S 1 = S 1 × R. Let M be a manifold and let ω be an “angular variable,” that is a smooth map ω : M → S 1 . Define dω, a one-form on M by taking the R-projection of T ω. Show that (i) if ω : M → S 1 , then ddω = 0; and (ii) if f : M → N is smooth, then f ∗ (dω) = d(f ∗ ω), where f ∗ ω = ω ◦ f .  7.4-10.

Prove the identity £X iY − £Y iX − i[X,Y ] = [d, iX ◦ iY ]. 1

 7.4-11. (i) Let X = (X , X , 0) be a vector field defined on the plane S = { (x, y, 0) | x, y ∈ R } in R3 . Show that there exists Y ∈ X(R3 ) such that X = curl Y on S. Hint: Let

2

Y (x, y, z) = (zX 2 (x, y), −zX 1 (x, y), 0).

(ii) Let S be a closed surface on R3 and X ∈ X(S). Show that there exists Y ∈ X(R3 ) such that X = curl Y on S. ˜ ∈ X(R3 ) and put ω = ∗X ˜ [ . Locally find α such that dα = ω Hint: By Theorem 6.5.9 extend X to X P P by (i). Use a partition of unity (ϕi ) to write ω = ϕi ω and let dαi = ϕi ω, α = αi .

434

7. Differential Forms

(iii) Generalize this to forms on a closed submanifold of a manifold admitting C k -partitions of unity.  7.4-12. Let M be a manifold and α ∈ Ωk (M ). If τM : T M → M denotes the tangent bundle projection, let α0 = τ ∗ α ∈ Ωk (T M ). A k-form Γ on T M for which there is an α ∈ Ωk (m) such that α0 = Γ is called basic. A vector field X ∈ X(T M ) is said to be vertical if T τM ◦ X = 0. Show that Γ ∈ Ωk (T M ) is basic if and only if iX Γ = 0, and £X Γ = 0 for any vertical vector field χ on T M . Conclude that if Γ is closed it is basic if and only if iX Γ = 0 for every vertical vector field X on T M . Hint: Since X and the zero vector field on M are τM -related, if Γ is vertical, the two identities follow. Conversely, if Ft is the flow of X, then Ft∗ Γ = Γ. Define α ∈ Ωk (T M ) by α(m)(v1 , . . . , vk ) = Γ(u)(V1 , . . . , Vk ), where τ (u) = m, Tm τM (Vi ) = vi , i = 1, . . . , k. Show that this definition is independent of the choices of u, V1 , . . . , Vk in the following way. Let τ (u0 ) = m, Tm τM (Vi0 ) = vi , i = 1, . . . , k, w = u − u0 . Consider the local flow Ft in a vector bundle chart of T M containing Tm M which occurs only in the fibers and which on Tm M itself is translation by tw. The vector field it generates is vertical so that Tt∗ Γ = Γ and F1 (u) = u0 . Let Tu0 F1 (Vi0 ) = Vi00 ∈ Tu (T M ) and show Tu τ (Vi00 ) = vi since τ ◦ ϕt = τ ; thatis, Vi00 − Vi is a vertical vector. Now use the fact that Vi00 − Vi contracts with Γ to give zero to prove inductively that Γ(u)(Vi , . . . , Vk ) = Γ(u0 )(Vi0 , . . . , Vk0 ).  7.4-13.

Show that on R4 , the ideal generated by ω1 = x2 dx1 + x3 dx4 ,

ω2 = x3 dx2 + x2 dx3

is a differential ideal. Find its integral manifolds.

7.5

Orientation, Volume Elements and the Codifferential

Orientation and Volume Manifolds. This section globalizes the setting of §7.2 from linear spaces to manifolds. All manifolds in this section are finite dimensional.2 7.5.1 Definition. A volume form on an n-manifold M is an n-form µ ∈ Ωn (M ) such that µ(m) 6= 0 for all m ∈ M ; M is called orientable if there exists some volume form on M . The pair (M, µ) is called a volume manifold. Thus, µ assigns an orientation, as defined in equation (7.2.5), to each fiber of T M . For example, R3 has the same standard volume form µ = dx ∧ dy ∧ dz. 7.5.2 Proposition.

Let M be an n-manifold.

(i) M is orientable iff there is an element µ ∈ Ωn (M ) such that every other ν ∈ Ωn (M ) may be written ν = f µ for some f ∈ F(M ). (ii) If M is orientable then it has an atlas {(Ui , ϕi )}, where ϕi : Ui → Ui0 ⊂ Rn , such that the Jacobian determinant of the overlap maps is positive (the Jacobian determinant is the determinant of the derivative, a linear map from Rn into Rn ). The converse is true if M is paracompact. Proof. For (i) assume first Vn that M is orientable, with a volume form µ. Let ν be any other element of Ωn (M ). Now each fiber of (M ) is one-dimensional, so we may define a map f : M → R by ν 0 (m) = f (m)µ0 (m) 2 For

where µ(m) = µ0 (m)dx1 ∧ · · · ∧ dxn

infinite-dimensional analogues of orientability, see Elworthy and Tromba [1970b].

7.5 Orientation, Volume Elements and the Codifferential

435

and similarly for ν 0 . Since, µ0 (m) 6= 0 for all m ∈ M , F (m) = ν 0 (m)/µ0 (m) is of class C ∞ . Conversely, if Ωn (M ) is generated by ν, then ν 0 (m) 6= 0 for all m ∈ M since each fiber is one-dimensional. To prove (ii), let {(Ui , ϕi )} be an atlas with ϕi (Ui ) = Ui0 ⊂ Rn . We may assume that all Ui0 are connected by taking restrictions if necessary. Now (ϕi )∗ µ = fi dx1 ∧ · · · ∧ dxn = fi µ0 , where µ0 is the standard volume form on Rn . By means of a reflection if necessary, we may assume that fi (u0 ) > 0 (fi 6= 0 since ν is a volume form). However, a continuous real-valued function on a connected space that is nowhere zero is always > 0 or always < 0. Hence, for overlap maps we have

ϕi ◦ ϕ−1 dx1 ∧ · · · ∧ dxn = ϕi∗ ◦ ϕ−1 dx1 ∧ · · · ∧ dxn j j ∗ ∗ =

fi dx1 ∧ · · · ∧ dxn . fj ◦ ϕj ◦ ϕ−1 i

But ψ ∗ (u)(α1 ∧ · · · ∧ αn ) = Dψ(u)∗ · α1 ∧ Dψ(u)∗ · α2 ∧ · · · ∧ Dψ(u)∗ · αn . where Dψ(u)∗ · α1 (e) = α1 (Dψ(u) · e). Hence, by the definition of determinant, det(D(ϕj ◦ ϕ−1 i )) =

fi (u) > 0. fj [(ϕj ◦ ϕ−1 i )(u)]

(We leave as an exercise the fact that the canonical isomorphism L(E, E) ≈ L(E ∗ , E ∗ ) used before does not affect determinants.) For the converse of (ii), suppose {(Vα , πα )} is an atlas with the given property, and let {(Ui , ϕi , gi )} a subordinate partition of unity. Let µi = ϕ∗i (dx1 ∧ · · · ∧ dxn ) ∈ Ωn (Ui ) n and let P µ˜i (m) = gi (m)µi (m) if m ∈ Ui and µ˜i = 0 if 6∈ Ui . Since supp(gi ) ⊂ Ui , we have µ˜i ∈ Ω (M ). Let µ = i µ˜i . Since this sum is finite in some neighborhood point, it is clear from local representatives that µ ∈ Ωn (M ). To show that µ is a volume form on M , notice that on Ui ∩ Uj 6= ∅ we have ∗ 1 n µj = ϕ∗j (dx1 ∧ · · · ∧ dxn ) = ϕ∗i (ϕj ◦ ϕ−1 i ) (dx ∧ · · · ∧ dx ) ∗ 1 n = [det D(ϕj ◦ ϕ−1 i ) ◦ ϕi ] ◦ ϕi (dx ∧ · · · ∧ dx ) = [det D(ϕj ◦ ϕ−1 i ) ◦ ϕi ]µi = aji µi

where aji ∈ F(Ui ∩ Uj ), aji > 0 and there is no implied sum. By local finiteness of the covering {Ui }, a given point m ∈ M belongs only to a finite number of open sets, say Ui0 , Ui1 , . . . , UiN . Thus, (N ) N X X µ(M ) = µik (m) = (1 + aik i0 (m)) µi0 (m) 6= 0 k=0

k=1

since µi0 (m) 6= 0 and each aik i0 (m) > 0. It follows that µ(m) 6= 0 for each m ∈ M .



Thus, if (M, µ) is a volume manifold we get a map from Ωn (M ) to F(M ); namely, for each ν ∈ Ωn (M ), there is a unique f ∈ F(M ) such that ν = f µ. 7.5.3 Definition. Let M be an orientable manifold. Two volume forms µ1 and µ2 on M are called equivalent if there is an f ∈ F(M ) with f (m) > 0 for all m ∈ M such that µ1 = f µ2 . (This is clearly an equivalence relation.) An orientation of M is an equivalence class [µ] of volume forms on M . An oriented manifold (M, [µ]), is an orientable manifold M together with an orientation [µ] on M . If [µ] is an orientation of M , then [−µ], (which is clearly another orientation) is called the reverse orientation.

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7. Differential Forms

The next proposition tells us when [µ] and [−µ] are the only two orientations. 7.5.4 Proposition. tations.

Let M be an orientable manifold. Then M is connected iff M has exactly two orien-

Proof. Suppose M is connected, and µ, ν are two volume forms with ν = f µ. Since M is connected, and f (m) 6= 0 for all m ∈ M , either f (m) > 0 for all m or else f (m) < 0 for all m. Thus ν is equivalent to µ or −µ. Conversely, if M is not connected, let U (not equal to either ∅ or M ), be a subset that is both open and closed. If ν is a volume form on M , define χ by letting χ(m) = ν(m) if m ∈ U and χ(m) = −ν(m) if m 6∈ U . Obviously, χ is a volume form on M , and ν 6∈ [ν] ∪ [−ν].  A simple example of a nonorientable manifold is the M¨obius band (see Figure 7.5.1 and Exercise 7.5-12), For other examples, see Exercises 7.5-11 and 7.5-13.

Figure 7.5.1.

7.5.5 Proposition. The equivalence relation in Definition 7.5.3 is natural with respect to mappings and diffeomorphisms. That is, if f : M → N is of class C ∞ , µN and νN are equivalent volume forms on N , and f ∗ (µN ) is a volume form on M , then f ∗ (νN ) is an equivalent volume form. If f is a diffeomorphism and µM and νM are equivalent volume forms on M , then f∗ (µM ) and f∗ (νM ) are equivalent volume forms on N. Proof. This follows from the fact that f ∗ (gω) = (g ◦ f )f ∗ ω, which implies f∗ (gω) = (g ◦ f −1 )f∗ ω when f is a diffeomorphism.  7.5.6 Definition. Let M be an orientable manifold with orientation [µ]. A chart (U, ϕ) with ϕ(U ) = U 0 ⊂ Rn is called positively oriented if ϕ∗ (µ|U ) is equivalent to the standard volume form dx1 ∧ · · · ∧ dxn ∈ Ωn (U 0 ). From Proposition 7.5.5 we see that this definition does not depend on the choice of the representative from [µ]. If M is orientable, we can find an atlas in which every chart has positive orientation by choosing an atlas of connected charts and, if a chart has negative orientation, by composing it with a reflection. Thus, in Proposition 7.5.2(ii) the atlas consists of positively oriented charts. ˜ and a two-toOrientable Double Covering. If M is not orientable, there is an orientable manifold M ∞ ˜ ˜ one C surjective local diffeomorphism π : M → M . The manifold M is called the orientable double covering and is useful for reducing certain facts to the orientable case. The double covering is constructed as follows. Let ˜ = { (m, [µm ]) | m ∈ M, [µm ] an orientation of Tm M }. M

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Define a chart at (m, [µm ]) in the following way. Fix an orientation [ω] of Rn and an orientation reversing isomorphism A of Rn , for example, the isomorphism given by A(e1 ) = −e1 , A(ei ) = ei , i = 2, . . . , n where [ei , . . . , en ] is the standard basis of Rn . If ϕ : U → U 0 ⊂ Rn is a chart of M at m, setting U ± = { (u, [µu ]) | u ∈ U, [ϕ∗ (µu )] = [±ω] }, and defining ϕ± → U 0 by ϕ+ (u, [µu ]) = ϕ(u),

ϕ− (u, [µu ]) = (A ◦ ϕ)(u),

˜ . It is straightforward to check that the family {(U ± , ϕ± )} constructed in this we get charts (U ± , ϕ± ) of M ˜ into a differentiable n-manifold. Define π : M ˜ → M by π(m, [µm ]) = m. way forms an atlas, thus making M In local charts, π is the identity mapping, so that π is a surjective local diffeomorphism. Moreover π −1 (m) = {(m, [µm ]), (m, [−µm ])}, ˜ is orientable, since the atlas formed by the charts (U ± , ϕ± ) so that π is a twofold covering of M . Finally, M ˜ at the point (m, [µm ]) by is orientable. A natural orientation on M is induced on the tangent space to M ∗ [(Tm π) µm ]. ˜ is connected iff M is nonorientable. In 7.5.7 Proposition. Let M be a connected n-manifold. Then M ˜ fact, M is orientable iff M consists of two disjoint copies of M , one with the given orientation, the other with the reverse orientation. Proof. The if part of the second statement is a reformulation of Proposition 7.5.4 and it also proves that if ˜ is connected, then M is nonorientable. Conversely if M is a connected manifold and if M ˜ is disconnected, M ˜ let C be a connected component of M . Then since π is a local diffeomorphism, π(C) is open in M . We shall ˜ be such that π(m prove that π(C) is closed. Indeed, if m ∈ cl(π(C)), let m ˜ 1, m ˜2 ∈ M ˜ 1 ) = π(m ˜ 2 ) = m. ˜ ˜ ˜ ˜ ∩ C = ∅, then shrinking U ˜1 If there exists neighborhoods U1 , U2 , of m ˜ 1, m ˜ 2 such that U1 ∩ C = ∅ and U ˜ ˜ ˜ and U2 if necessary, the open neighborhoods π(U1 ) and π(U2 ) of m have empty intersection with π(C), contradicting the fact that m ∈ cl(π(C)). Thus at least one of m ˜ 1, m ˜ 2 is in cl(C) = C; that is, m ∈ π(C) and ˜ can have hence π(C) is closed. Since M is connected, π(C) = M . But π is double covering of M so that M at most two components, each of them being diffeomorphic to M . Hence M is orientable, the orientation being induced from one of the connected components via π.  Conditions for Orientability.

Another criterion of orientability is the following.

7.5.8 Proposition. Suppose M is an orientable n-manifold and V is a submanifold of codimension k with trivial normal bundle. That is, there are C ∞ maps Ni : V → T M , i = 1, . . . , k such that Ni (v) ∈ Tv (M ), and Ni (v) span a subspace Wv , such that Tv M = Tv V ⊕ Wv for all v ∈ V . Then V is orientable. Proof. Let µ be a volume form on M . Consider the restriction µ|V : V → Γn (M ). Let us first note that µ|Γ is a smooth mapping of manifolds. This follows by using charts with the submanifold property, where the local representation is a restriction to a subspace. Now define µ0 : V → Γn−k (V ) as follows. For X1 , . . . , Xn−k ∈ X(V ), put µ0 (v)(X1 (v), . . . , Xn−k (v)) = µ(v)(N1 (v), . . . , Nk (v), X1 (v), . . . , Xn−k (v)). It is clear that µ0 (v) 6= 0 for all v. It remains only to show that µ0 is smooth, but this follows from the fact that µ|V is smooth.  If g is a Riemannian metric, then g [ : T M → T ∗ M denotes the index-lowering operator and we write g = (g [ )−1 . For f ∈ F(M ), grad f = g # (df ) #

is called the gradient of f . Thus, grad f ∈ X(M ). In local coordinates, if [gij ] = [g(∂/∂xi , ∂/∂xj )] and [g ij ] is the inverse matrix, then ∂f (7.5.1) (grad f )i = g ij j . ∂x

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7.5.9 Corollary. Suppose M is an orientable Riemannian manifold, H ∈ F(M ) and c ∈ R is a regular value of H. Then V = H −1 (c) is an orientable submanifold of M and, assuming it is nonempty, has codimension one. Proof. Suppose c is a regular value of H and H −1 (c) = V 6= ∅. Let N = grad H|V , the gradient of H restricted to V . Then N (v) 6∈ Tv V for v ∈ V , because Tv V is the kernel of dH(v)[N (v)] = g(N, N )(v) > 0 as dH(v) 6= 0 by hypothesis. Then Proposition 7.5.8 applies, and so V is orientable.



Thus if we interpret V as the “energy surface,” we see that it is an oriented submanifold for “almost all” energy values by Sard’s theorem. Orientation Preserving Maps. can now be defined.

The notion of orientation preserving maps between oriented manifolds

7.5.10 Definition. Let M and N be two orientable n-manifolds with volume forms µM and µN , respectively. Then we call a C ∞ map f : M → N volume preserving (with respect to µM and µN ) if f ∗ µN = µM , orientation preserving if f ∗ (µN ) ∈ [µM ], and orientation reversing if f ∗ (µN ) ∈ [−µM ]. An orientable manifold admitting (respectively, not admitting) an orientation reversing diffeomorphism is called reversible (respectively, non-reversible). From Proposition 7.5.5, [f ∗ µN ] depends only on [µN ]. Thus the first part of the definition explicitly depends on µM and µN , while the last four parts depend only on the orientations [µM ] and [µN ]. Furthermore, we see from Proposition 7.5.5 that if f is volume preserving with respect to µM and µN , then f is volume preserving with respect to hµM and gµN iff h = g ◦ f . It is also clear that if f is volume preserving with respect to µM and µN , then f is orientation preserving with respect to [µM ] and [µN ]. 7.5.11 Proposition. Let M and N be n-manifolds with volume forms µM and µN , respectively. Suppose f : M → N is of class C ∞ . Then f ∗ (µN ) is a volume form iff f is a local diffeomorphism; that is, for each m ∈ M , there is a neighborhood V of m such that f |V : V → f (V ) is a diffeomorphism. If M is connected, then f is a local diffeomorphism iff f is orientation preserving or orientation reversing. Proof. If f is a local diffeomorphism, then clearly F ∗ (µN )(m) 6= 0, by Proposition 7.2.3(ii). Conversely, if f ∗ (µM ) is a volume form, then the determinant of the derivative of the local representative is not zero, and hence the derivative is an isomorphism. The result then follows by the inverse function theorem. The second statement follows at once from the first and Proposition 7.5.4.  Jacobian Determinant.

Next we consider the global analog of the determinant.

7.5.12 Definition. Suppose M and N are n-manifolds with volume forms µM and µN , respectively. If f : M → N is of class C ∞ , the unique C ∞ function J(µM , µN )f ∈ F(M ) such that f ∗ µN = (J(µM , µN )f )µM is called the Jacobian determinant of f (with respect to µM and µN ). If f : M → M we write Jµ f = J(µ, µ)f . Note that J(µM , µN )f (m) = det(Tm f ), the determinant being taken with respect to the volume forms µM (m) on Tm M and µN (f (m)) on Tf (m) N . The basic properties of determinants that were developed in §7.2 also hold in the global case, as follows. First, we have the following consequences of Proposition 7.5.11. 7.5.13 Proposition. The C k map f : M → N , k ≥ 1, is a local C k diffeomorphism iff J(µM , µN )f (m) 6= 0 for all m ∈ M . Second, we have consequences of the definition and properties of pull-back. 7.5.14 Proposition.

Let (M, µ) be a volume manifold.

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(i) If f : M → M and g : M → M are of class C k , k ≥ 1, then Jµ (f ◦ g) = [(Jµ f ) ◦ g][Jµ g]. (ii) If h : M → M is the identity, then Jµ h = 1. (iii) If f : M → M is a diffeomorphism, then Jµ (f −1 ) = Proof.

1 . [(Jµ f ) ◦ f −1 ]

For (i), Jµ (f ◦ g)µ = (f ◦ g)∗ µ = g ∗ f ∗ µ = g ∗ (Jµ f )µ = ((Jµ f ) ◦ g)g ∗ µ = ((Jµ f ) ◦ g)(Jµ g)µ.

Part (ii) follows since h∗ is the identity. For (iii) we have Jµ (f ◦ f −1 ) = 1 = ((Jµf ) ◦ f −1 )(Jµ f −1 ).



7.5.15 Proposition. Let (M, [µM ]) and (N, [µN ]) be oriented manifolds and f : M → N be a map of class C k , k ≥ 1. Then f is orientation preserving iff J(µM , µN )f (m) > 0 for all m ∈ M , and orientation reversing if J(µM , µN )f (m) < 0 for all m ∈ M . Also, f is volume preserving with respect to µM and µN iff J(µM , µN )f = 1. This proposition follows from the definitions. Note that the first two assertions depend only on the orientations [µM ] and [µN ], since   g◦f J(hµM , gµN )f = J(µM , µN )f, h which the reader can easily check. Here g ∈ F(N ), h ∈ F(M ), g(n) 6= 0, and h(m) 6= 0 for all n ∈ N , m ∈ M. Divergence. We have seen that in R3 the divergence of a vector field is expressible in terms of the standard volume element µ = dx ∧ dy ∧ dz by the use of the metric in R3 (see Example 7.4.3D). There is, however, a second characterization of the divergence that does not require a metric but only a volume form µ, namely £F µ = (div F )µ, as a simple computation shows. This can now be generalized. 7.5.16 Definition. Let (M, µ) be a volume manifold; that is, M is an orientable manifold with a volume form µ. Let X be a vector field on M . The unique function divµ X ∈ F(M ), such that £X µ = (divµ X)µ is called the divergence of X. We say X is incompressible or divergence free (with respect to µ) if divµ X = 0. 7.5.17 Proposition.

Let (M, µ) be a volume manifold and X a vector field on M .

(i) If f ∈ F(M ) and f (m) 6= 0 for all m ∈ M , then divf µ X = divµ X +

£X f . f

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(ii) For g ∈ F(M ), divµ gX = g divµ X + £X g. Proof.

Since £X is a derivation, £X (f µ) = (£X f )µ + f £X µ.

As f µ is a volume form, (divf µ X)(f µ) = (£X f )µ + f (divµ X)µ. Then (i) follows. For (ii), £gX µ = g£X µ + dg ∧ iX µ, and from the antiderivation property of iX , dg ∧ iX µ = −iX (dg ∧ µ) + iX dg ∧ µ. But dg ∧ µ ∈ Ωn+1 (M ), and hence dg ∧ µ = 0. Also, iX dg = £X g, so £gX µ = g£X µ + (£X g)µ. The result follows from this.



7.5.18 Proposition. Let (M, µ) be a volume manifold and X a vector field on M . Then X is incompressible (with respect to µ) iff the flow of X is volume preserving; that is, the local diffeomorphism Ft : U → V is volume preserving with respect to µ|U and µ|V . Proof. Since X is incompressible, £X µ = 0, and so µ is constant along the flow of X; µ(m) = (Ft∗ µ)(m).  Thus Ft is volume preserving. Conversely, if (Ft∗ µ)(m) = µ(m), then £X µ = 0. 7.5.19 Corollary. Let (M, µ) be a volume manifold and X a vector field with flow Ft on M . Then X is incompressible iff Jµ Ft = 1 for all t ∈ R. One-Densities. The above developments regarding the Jacobian and divergence can also be carried out for one-densities. If |µM |, |µN | are one-densities and f : M → N is C ∞ , we shall write f ∗ |µN | = J(|µM |, |µN |, f )|µM |, where the pull back is defined as for forms. Then Propositions 7.5.13 and 7.5.14 go through for one-densities. The Lie derivative of a one-density is defined by d £X |µ| = F ∗ |µ| dt t=0 t and one defines the divergence of X with respect to |µ| as in Definition 7.5.16. Then it is easy to check that Proposition 7.5.17–Corollary 7.5.19 have analogues for one-densities. Riemannian Volume Forms. We shall now globalize the concepts pertaining to Riemannian volume forms and densities, as well as the Hodge star operator discussed in §7.2. 7.5.20 Proposition. Let (M, g) be a pseudo-Riemannian manifold of signature s; that is, g(m) has signature s for all m ∈ M . (i) If M is orientable, then there exists a unique volume element µ = µ(g) on M , called the g-volume (or pseudo-Riemannian volume of g ), such that µ equals one on all positively oriented orthonormal bases on the tangent spaces to M . If X1 , . . . , Xn is such a basis in an open set U of M and if ζ 1 , . . . , ζ n is the dual basis, then µ = ζ 1 ∧ · · · ∧ ζ n . More generally, if v1 , . . . , vn ∈ Tx M are positively oriented, then µ(x)(v1 , . . . , vn ) = | det[g(x)(vi , vj )]|1/2 .

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(ii) For every α ∈ R there exists a unique α-density |µ|α , called the g-α-density (or the pseudoRiemannian α-density of g ), such that |µ|α equals 1 on all orthonormal bases of the tangent spaces to M . If Xi , . . . , Xn is such a basis in an open set U of M with dual basis ζ 1 , . . . , ζ n , then |µ|α = |ζ 1 ∧ · · · ∧ ζ n |α . More generally, if v1 , . . . , vn ∈ Tx M , then |µ|α (x)(v1 , . . . , vn ) = | det[g(x)(vi , vj )]|α/2 . This is a consequence of Proposition 7.2.10 and the fact that µ and |µ|α are smooth. Also note that in an oriented chart (x1 , . . . , xn ) on M , we have µ = | det[gij ]|1/2 dx1 ∧ · · · ∧ dxn . As in the vector space situation, g induces a pseudo-Riemannian metric on Γk (M ) by hα, βix = αi1 ...ik β i1 ...ik , (7.5.2) Vk where the sum is over i1 < · · · < ik and where, α, β ∈ (M )x and β i1 ...ik are the components of the associated contravariant tensor to β. As in Proposition 7.2.11, if X1 , . . . Xn is an orthonormal basis in U ⊂ M with dual basis ζ 1 , . . . ζ n , then the elements ζ i1 ∧ · · · ∧ ζ ik , where i1 < · · · < ik form an orthonormal basis of Γk (U ). The Hodge Star Operator. On an orientable pseudo-Riemannian manifold with pseudo-Riemannian volume form µ, the Hodge operator is defined pointwise by ∗ : Ωk (M ) → Ωn−k (M ),

(∗α)(x) = ∗α(x);

that is, α ∧ ∗β = hα, βi µ for α, β ∈ Ωk (M ). The properties in Propositions 7.2.12 and 7.2.13 carry over since they hold pointwise. One can check that if α is C r then so is ∗α. The Codifferential. The exterior derivative and the Hodge star operator enable us to introduce the following linear operator δ. (The reason for the strange-looking factor (−1) in the definition is so a later integration by parts formula, proved in Corollary 8.2.13, will come out simple.) 7.5.21 Definition. forms β by

The codifferential δ : Ωk+1 (M ) → Ωk (M ) is defined by δ(Ω0 (M )) = 0 and on k + 1 δβ = (−1)nk+1+Ind(g) ∗d∗β.

Notice that since d2 = 0 and ∗∗ is a multiple of the identity, δ 2 = 0. For example, in R3 , let α = ady ∧ dz − bdx ∧ dz + cdx ∧ dy. Then ∗α = adx + bdy + cdz, so d∗α = (bx − ay )dx ∧ dy + (cx − az )dx ∧ dz + (cy − bz )dy ∧ dz and ∗d∗α = (bx − ay )dz − (cc − az )dy + (cy − bz )dx. Thus, as nk + 1 + Ind (g) = 4 is even, dα = (cγ − bz )dx + (az − cx )dy + (bx − ay )dz. The formula for δβ in coordinates is given by 1 | det[gij ]|−1/2 gi1 r1 . . . gik rk k+1 ! k+1 X (−1)p g r1 j1 · · · g rp−1 jp6=1 g ljr g rp ip6=1 βj1 ...jk6=1 | det[gij ]|1/2

(δβ)i1 ...ik = ×

∂ ∂xl

p=1

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or as a contravariant tensor (δβ)r1 ...rk =

1 | det[gij ]|−1/2 k+1 ! k+1 X ∂ p r1 ...rp−1 lrp ...rk 1/2 × l (−1) β | det[gij ]| , ∂x p=1

where β = βr1 ...rk rk+1 dxr1 ∧ · · · ∧ dxrk+1 , (sum over r1 < · · · < rk+1 ) is the usual coordinate expression for β. The coordinate formula is messy to prove directly. However it follows fairly readily from integration by parts in local coordinates and that fact that δ is the adjoint of d, a fact that will be proved in Chapter 7 (see Corollary 8.2.13 and Exercise 8.5-7). We now express the divergence of a vector field X ∈ X(M ) in terms of δ. We define the divergence divg (X) with respect to a pseudo-Riemannian metric g to be the divergence of X with respect to the pseudo-Riemannian volume µ = µ(g); that is, £X µ = divg (X)µ. To compute the divergence, we prepare a lemma. 7.5.22 Lemma. Proof.

iX µ = ∗X [ .

Let v2 , . . . , vn ∈ Tx M be orthonormal and orthogonal to X(n). From Proposition 7.2.10, we have iX µ(v2 , . . . , vn ) = µ(X(x), v2 , . . . , vn ) p = g(X(x), X(x)).

On the other hand, we claim that ∗X [ =

p g(X(x), X(x)) v2[ ∧ · · · ∧ vn[

Indeed, this may be verified using the definition in Proposition 7.2.12 with x a 1-form and β = X [ . Using this formula for ∗X [ , we get p ∗X [ (x)(v2 , . . . , vn ) = g(X(x), X(n)) = iX µ(v2 , . . . , vn ). Equality on such v2 , . . . , vn implies equality, as is readily seen.



This may be seen in coordinates using formula (7.2.12). 7.5.23 Proposition.

Let g be a pseudo-Riemannian metric on the orientable n-manifold M . Then divg (X) = −δX [

(7.5.3)

In (positively oriented ) local coordinates divg (X) = | det[g ij ]|−1/2 Proof.

 ∂  1/2 k | det[g ]| X ij ∂xk

(7.5.4)

Let £X µ = diX µ = d∗X [

by the lemma. But then (divg X)µ = −∗δX [ = −(δX [ )∗1 by the definition of δ and the formula for ∗∗. Since ∗1 = µ, we get equation (7.5.3). To prove formula (7.5.4), write µ = | det[gij ]|1/2 dx1 ∧ · · · ∧ dxn and compute £X µ = diX µ in these coordinates. We have iX µ = det[gij ]1/2 X k (−1)k dx1 ∧ · · · ∧ dxk ∧ · · · ∧ dxn

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and so 

 ∂ 1/2 k | det[g ]| X dx1 ∧ · · · ∧ dxn . ij ∂xk   1 1 ∂/∂xk = | det[g ]| X µ. ij k 2 | det[gij ]|1/2

diX µ =



7.5.24 Definition. The Laplace–Beltrami operator on functions on an orientable pseudo-Riemannian manifold is defined by ∇2 = div ◦ grad. Thus, in a positively oriented chart, equation (7.5.4) gives   ∂f ∂ ∇2 f = | det gij |−1/2 k g lk | det gij |1/2 l . ∂x ∂x

(7.5.5)

Notice that this equation gives a formula for the Laplace-Beltrami operator in an arbitrary coordinate system; the reader has probably seen the Laplacian in, say, spherical coordinates—such a formula is consistent with the one given here. Haar measure. We conclude this section with a link to Lie groups. One can characterize Lebesgue measure up to a multiplicative constant on Rn by its invariance under translations. Similarly, on a locally compact group there is a unique (up to a nonzero multiplicative constant) left-invariant measure, called Haar measure. For Lie groups the existence of such measures is especially simple. 7.5.25 Proposition. Let G be a Lie group. Then there is a volume form µ, unique up to nonzero multiplicative constants, that is left invariant. If G is compact, µ is right invariant as well. Proof.

Pick any n-form µe on Te G that is nonzero and define an n-form on Tg G by µg (v1 , . . . , vn ) = µe · (T Lg−1 v1 , . . . , T Lg−1 · vn ).

Then µg is left invariant and smooth. For n = dim G, µe is unique up to a scalar factor, so µg is as well. Fix g0 ∈ G and consider Rg∗0 µ = cµ for a constant c. If G is compact, this relationship may be integrated, and by the change of variables formula we deduce that c = 1. Hence, µ is also right invariant. 

Exercises  7.5-1. Let f : Rn → Rn be a diffeomorphism with positive Jacobian and f (0) = 0. Prove that there is a continuous curve ft of diffeomorphisms joining f to the identity. Hint: First join f to Df (0) by gt (x) = f (tx)/t.  7.5-2.

If t is a tensor density of M , that is, t = t0 ⊗ µ, where µ is a volume form, show that £X t = (£X t0 ) ⊗ µ + (divµ X)t0 ⊗ µ.

 7.5-3. Let E be a Banach space. A map A : E → E is said to be derived from a variational principle if there is a function L : E → R such that dL(x)v = hA(x), vi, where h , i is an inner product on E. Prove Vainberg’s theorem: A comes from a variational principle if and only if DA(x) is a symmetric linear operator. do this by applying the Poincar´e lemma to the one-form α(x)v˙ = hA(x), vi (see Marsden and Hughes [1983]).

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 7.5-4. Show in three different ways that the sphere S n is orientable by using Proposition 7.5.2 and the two charts given in Figure 3.1.2, by constructing an explicit n-form, and by using Corollary 7.5.9.  7.5-5.

Use formula (7.5.5) to show that in polar coordinates (r, θ) in R2 , ∇2 f =

∂2f 1 ∂2f 1 ∂f + 2 2 + , 2 ∂r r ∂θ r ∂r

and that in spherical coordinates (ρ, θ, φ) in R3 ,   ∂2f ∂ ∂f 1 ∂2f +ρ 2, ∇2 f = (1 − µ2 ) + 2 2 ∂µ ∂µ 1 − µ ∂θ ∂ρ where µ = cos φ.  7.5-6.

Let (M, µ) be a volume manifold. Prove the identity divµ [X, Y ] = X[divµ Y ] − Y [divµ X].

 7.5-7. Let f : M → N be a diffeomorphism of connected oriented manifolds with boundary. Assuming that Tm f : Tm M → Tf (m) N is orientation preserving for some m ∈ Int(M ), show that J(f ) > 0 on M ; that is, f is orientation preserving.  7.5-8. Let g be a pseudo-Riemannian metric on M and define gλ = λg for λ > 0. Let ∗λ be the Hodge-star operator defined by gλ and set ∗1 = ∗. Show that if λ ∈ Ωk (M ), then ∗λ α = λ(n/2)−k ∗α.  7.5-9. In R3 equipped with the standard Euclidean metric, show that for any vector field F and any function f we have: div F = −δF [ , curl F = (δ∗F [ )# , and grad f = −(∗δ∗f )# .  7.5-10.

Show that if M and N are orientable, then so is M × N .

 7.5-11. (i) Let σ : M → M be an involution of M , that is, σ ◦ σ = identity, and assume that the equivalence relation defined by σ is regular, that is, there exists a surjective submersion π : M → N such that π −1 (n) = {m, σ(m)}, where π(m) = n. Let Ω± (M ) = { α ∈ Ω(M ) | σ ∗ α = ±α } be the ±1 eigenspaces of σ ∗ . Show that π ∗ : Ω(N ) → Ω+ (M ) is an isomorphism. Hint: To show that range π ∗ = Ω+ (M ), note that π ◦ σ = π implies range(π ∗ ) ⊂ Ω+ (M ). For the converse conclusion, note that Tm π is an isomorphism, so a form at π(m) uniquely determines a form at m. Show that this resulting form is smooth by working in a chart on M diffeomorphic to a chart on N . (ii) Show that RPn is orientable if n is odd and is not orientable if n is even. Hint: In (i) take M = S n ⊂ Rn+1 , σ(x) = x, and N = RPn . Let ω be a volume element on S n induced by a volume element of Rn+1 . Show that σ ∗ ω = (−1)n+1 ω. Now apply (i) to orient RPn for n odd. If n is even, let ν be an n-form on RPn ; then π ∗ ν = f ω. Show that f (x) = −f (−x) so f must vanish at a point of S n .  7.5-12. In Example 3.4.10C, the M¨ obius band M was defined as the quotient of R2 by the equivalence k relation (x, y) ∼ (x + k, (−1) y) for any k ∈ Z. (i) Show that this equivalence relation is regular. Show that M is a non-compact, connected, two-manifold.

7.5 Orientation, Volume Elements and the Codifferential

445

(ii) Define the map σ : R2 → R2 by σ(x, y) = (x + 1, −y). Show that π ◦ σ = σ, where σ : R2 → M, is the canonical projection. If ν ∈ Ω2 (M) define f ∈ F(R2 ) by π ∗ ν = f ω, where ω is an area form on R2 . Show that f (x + 1, −y) = −f (x, y). (iii) Conclude that f must vanish at a point of R2 , and that this implies M is not orientable.  7.5-13.

The Klein bottle K is defined as the quotient of R2 by the equivalence relation defined by (x, y) ∼ (x + n, (−1)n y + m)

for any n, m ∈ Z. (i) Show that this equivalence relation is regular. Show that K is a compact, connected, smooth, twomanifold. (ii) Use Exercise 7.5-12(ii), (iii) to show that K is non-orientable.  7.5-14 (Orientation in vector bundles). Let π : E → B be a vector bundle with finite dimensional fiber modeled on a vector space E, and assume that B is connected. The vector bundle is said to be orientable if the line bundle L = E ∗ ∧ · · · ∧ E ∗ (dim(E) times) has a global nowhere vanishing section. An orientation of E is an equivalence class of global nowhere vanishing sections of L under the equivalence relation: σ1 ∼ σ2 iff there exists f ∈ F(B), f > 0 such that σ2 = f σ1 . (i) Prove that E is orientable iff L is a trivial line bundle. Show that E admits exactly two orientations. Show that an orientation [σ] of E induces an orientation in each fiber of E. (ii) Show that a manifold M is orientable iff its tangent bundle is an oriented vector bundle. (iii) Let E, F be vector bundles over the same base. Show that if two of E, F , and E ⊕ F are orientable, so is the third. (iv) Let E, F be vector bundles (over possibly different bases). Show that E × F is orientable if and only if E and F are both orientable. Conclude that if M, N are finite dimensional manifolds, then M × N is orientable if and only if M and N are both orientable. (v) Show that E ⊕ E ∗ is an orientable vector bundle if E is any vector bundle. Hint: Consider the section Ω(x)((e1 , α1 ), (e2 , α2 )) = hα2 , e1 i − hα1 , e2 i of (E ⊕ E ∗ ) ∧ (E ⊕ E ∗ ). (vi) Choose an orientation of the vector space E and assume B admits partitions of unity. Show that the vector bundle atlas all of whose change of coordinate maps have positive determinant relative to the orientation of E, when restricted to the fiber. Hint: If E is oriented and ψ : π −1 (U ) → U 0 × E is a vector bundle chart with U 0 open in the model space of B, and U is connected in B, define φ : π −1 (U ) → U 0 × F by φ(e) = ψ(e) if the linear map ψb : π −1 (b) → F is orientation preserving and φ(e) = (α ◦ ψ)(e), where α : F → F is an orientation reversing isomorphism of F , if ψb is orientation reversing. For the converse, choose a volume form ω on F and define on a vector bundle chart (V, φ) of E, with U connected in B, π −1 (U ) = V , ω(U ) : U → L|U by ω(U )(b)(e1 , . . . , er ) = ω(φ0 (b))(φb (e1 ), . . . , φb (er )), where r = dim F , b ∈ B, ei ∈ π −1 (b), i = 1, . . . , r, and φ0 : U → U 0 is the induced chart on B. Show that if b ∈ U1 ∩ U2 , then
ω(U1 )(b) = detω φ1b ◦ (φ2b )−1 ω(U2 ), where (π −1 (Ui ), φi ) are vector bundle charts, Ui connected, i = 1, 2. Next, glue the ω(U )’s together using a partition of unity.

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7. Differential Forms

(vii) Use (iv) to show that if E and F are oriented, then there exists a vector bundle atlas on E such that all φb : π −1 (b) → F are orientation preserving isomorphisms. Such an atlas is called positively oriented. (viii) Let E, F, B be finite dimensional, E oriented by σ and B oriented by ω. Show that π ∗ ω ∧ σ is a volume form on E. Conclude that an orientation of B and an orientation of F uniquely determine an orientation of E as a manifold. This orientation is called the local product orientation. (ix) Show that any vector bundle π : E → B with finite dimensional fiber has an oriented double cover ˜ → B, ˜ where π ˜:E ˜ = { (b, [µb ]) | b ∈ B, [µb ] is an orientation of π −1 (b) }, B ˜ → B is the map p(b, [µb ]) = b, and E ˜ = p∗ E. Find the vector bundle charts of E ˜ and show p : B ∗ ˜ that the fiber at (b, [µb ]) is oriented by [˜ p (µb )], where p˜ : E → E is the mapping induced by p on the ˜ If E = T B, what is E? ˜ pull-back bundle E.  7.5-15. Let M be a compact manifold and M the space of Riemannian metrics on M . Let T be a space of tensor fields on M of a fixed type. A mapping Φ : M → T is called covariant if for every diffeomorphism φ : M → M , we have Φ(φ∗ g) = φ∗ Φ(g). (i) Show that covariant maps satisfy the identity DΦ(g) · £X g = £X Φ(g) for every vector field X. (Assume Φ is differentiable and M, T are given suitable Banach space topologies.) (ii) Show that if M is oriented, then the map g 7→ µ(g), the volume element of g, is covariant. Is the identity in (i) anything interesting?  7.5-16. Let X be a vector field density on the oriented n-manifold M ; that is, X = F ⊗ µ, where F is a vector field and µ is a density. Use Exercise 7.5-2 to define div X and to show it makes intrinsic sense.  7.5-17. Show that an orientable line bundle over a base admitting partitions of unity is trivial. Hint: Since the bundle is orientable there exist local charts which when restricted to each fiber give positive functions. Regard these as local sections and then glue.  7.5-18. Let π : E → S 1 be a vector bundle with n-dimensional fiber. If E is orientable show that it is isomorphic to a trivial bundle over S 1 . Show that if ρ : F → S 1 is a non-orientable vector bundle with n-dimensional fiber and if E is non-orientable, then E and F are isomorphic. Conclude that there are exactly two isomorphic classes of vector bundles with n-dimensional fiber over S 1 . Construct a representative for the class corresponding to the non-orientable case. Hint: Construct a non-orientable vector bundle like the M¨obius band: the equivalence relation has a factor (−1)2n−1 if the dimension of the fiber is 2n − 1 or 2n. To prove non-orientability, proceed as in Exercise 7.5-12. Let {e1 , . . . , en+1 } be the standard bases of Rn+1 and

 7.5-19.

Ωn+1 = e1 ∧ · · · ∧ en+1 n

be the induced volume form. On S define ωn ∈ Ωn (S n ) by ωn (s)(v1 , . . . , vn ) = Ωn+1 (s, v1 , . . . , vn ) n

n

for s ∈ S , v1 , . . . , vn ∈ Ts S . (i) Use Proposition 7.5.8 to show that ωn is a volume form on S n ; ωn is called the standard volume form on S n . (ii) Let f : R+ × Rn+1 \{0} be given by f (t, s) = ts, where R+ is defined to be the set { t ∈ R | t > 0 }. Show that if R+ is oriented by dt, S n by ωn , and Rn+1 by Ωn+1 , then (Jf )(t, s) = tn . Conclude that f is orientation preserving.

Page 447

8 Integration on Manifolds

This version: February 22, 2007

The integral of an n-form on an n-manifold is defined by piecing together integrals over sets in Rn using a partition of unity subordinate to an atlas of coordinate charts. The change of variables theorem guarantees that the integral is well defined, independent of the choice of atlas and partition of unity. Two basic theorems of integral calculus, the change of variables theorem and Stokes’ theorem, are discussed in detail along with some applications. An important ingredient, which will be defined in this chapter is the notion of a manifold M with boundary ∂M . This generalizes the usual notion of a region or a surface with boundary that one learns in vector calculus. The general Stokes’ theorem1 for an n-manifold M with boundary ∂M is written as Z Z dω = ω M

∂M

where ω is an (n − 1)-form on M . This result captures—in one elegant formula—the classical theorems of Green, Gauss and Stokes, all which then appear as special cases. Of course we should not forget that the Fundamental Theorem of Calculus is also a key special case (where M is an interval and ∂M consists of the two enpoints). This beautiful unification of the classical integral theorems of vector calculus was one of the motivating drivers for the development of the theory of differential forms by Cartan in the early 1900s and moreover, it plays a crucial role in subsequent links between topology and differential forms in the de Rham theory that we study briefly in §8.5.

8.1

The Definition of the Integral

The aim of this section is to define the integral of an n-form on an oriented n-manifold M and prove a few of its basic properties. We begin with a summary of the basic results in Rn . 1 It

is said that Stokes’ learned the theorem from William Thomson (Lord Kelvin) and so some people call it the KelvinStokes’ theorem; how the general theorem came to be called the “general Stokes’ theorem” and not, say the “general Green theorem” is not clear to us.

448

8. Integration on Manifolds

Integration on Rn . Suppose f : Rn → R is continuous and has compact support. Then defined to be the Riemann integral over any rectangle containing the support of f . 8.1.1 Definition. basis of Rn ,

R

f dx1 . . . dxn is

Let U ⊂ Rn be open and ω ∈ Ωn (U ) have compact support. If, relative to the standard ω(x) =

1 ωi ...i (x)dxi1 ∧ · · · ∧ dxin = ω1...n (x)dx1 ∧ · · · ∧ dxn , n! 1 n

where the components of ω are given by ωi1 ...in (x) = ω(x)(ei1 , . . . , ein ), then we define Z

Z

ω1...n (x)dx1 · · · dxn .

ω= Rn

U

Recall now the change of variables theorem in Rn . Namely, if ζ is an integrable function and f : Rn → n R is a diffeomorphism, then ζ ◦ f is integrable and Z ζ(x1 , . . . , xn )dx1 · · · dxn Rn Z = |JΩ f (x1 , . . . , xn )|(ζ ◦ f )(x1 , . . . , xn )dx1 · · · dxn , (8.1.1) Rn

where Ω = dx1 ∧ · · · ∧ dxn is the standard volume form on Rn and JΩ f is the standard Jacobian determinant of f (and equals the Jacobian determinant relative to Ω.) This change of variables theorem can be rephrased in terms of pull backs in the following form. 8.1.2 Theorem (Change of Variables in Rn ). Let U and V be open subsets of Rn and suppose f : U → V is an orientation-preserving diffeomorphism. If ω ∈ Ωn (V ) has compact support, then f ∗ ω ∈ Ωn (U ) has compact support as well and Z Z f ∗ω = ω (8.1.2) U

V

Proof. If ω = ω1...n dx1 ∧ · · · ∧ dxn , then f ∗ ω = (ω1...n ◦ f )(JΩ f )Ω, where the n-form Ω = dx1 ∧ · · · ∧ dxn is the standard volume form on Rn . As discussed in §7.5, JΩ f > 0. Since f is a diffeomorphism, the support of f ∗ ω is f −1 (supp ω), which is compact. Then from equation (8.1.1), Z Z ∗ f ω= (ω1...n ◦ f )(JΩ f )dx1 · · · dxn U Rn Z Z = ω1...n dx1 · · · dxn = ω.  Rn

V

Integration on a Manifold. Suppose that (U, ϕ) is a chart on a manifold M and ω ∈ Ωn (M ) has compact support. If supp(ω) ⊂ U , we may form ω|U , which has the same support. Then ϕ∗ (ω|U ) has compact support, so we may state the following. 8.1.3 Definition. Let M be an orientable n-manifold with orientation [Ω]. Suppose ω ∈ Ωn (M ) has compact support C ⊂ U , where (U, ϕ) is a positively oriented chart. Then we define Z Z ω = ϕ∗ (ω|U ). (ϕ)

8.1 The Definition of the Integral

449

8.1.4 Proposition. Suppose ω ∈ Ωn (M ) has compact support C ⊂ U ∩ V , where (U, ϕ), and (V, ψ) are two positively oriented charts on the oriented manifold M . Then Z Z ω= ω. (ϕ)

Proof.

(ψ)

By Theorem 8.1.2, Z

Z

ψ ◦ ϕ−1

ϕ∗ (ω|U ) =


∗

ϕ∗ (ω|U ).

ψ∗ (ω|U ). (Recall that for diffeomorphisms, we have f∗ = (f −1 )∗ and (f ◦ g)∗ =  R R Thus, we may define U ω = (ϕ) ω, where (U, ϕ) is any positively oriented chart containing the compact R support of ω. More generally, we can define M ω where ω has compact support not necessarily lying in a single chart as follows.

Therefore, f∗ ◦ g∗ .)

R

ϕ∗ (ω|U ) =

R

8.1.5 Definition. Let M be an oriented manifold and A an atlas of positively oriented charts. Let P = {(Uα , ϕα , gα )} be a partition of unity subordinate to A. Define ωα = gα ω (so ωα has compact support in some Ui ) and let Z XZ ω= ωα . (8.1.3) P

α

(i) The sum in equation (8.1.3) contains only a finite number of nonzero terms.

8.1.6 Proposition.

(ii) For any other atlas of positively oriented charts and subordinate partition of unity Q we have Z Z ω= ω. P

The common value is denoted

R M

Q

ω, and is called the integral of ω ∈ Ωn (M ).

Proof. For any m ∈ M , there is a neighborhood U such that only a finite number of gα are nonzero on U . By compactness of supp ω, a finite number of such neighborhoods cover the support of ω. Hence only a finite number of gα are nonzero on the union of these U . For (ii), let P = {(Uα , ϕα , gα )} and Q = {(Vβ , ψβ , hβ )} be two partitions of unity with positively oriented charts. Then the functions {gα hβ } satisfy gα hβ (m) = 0 except for a finite number of indices (α, β), and Σα Σβ gα hβ (m) = 1, for all m ∈ M . Since Σβ hβ = 1, we get Z Z XZ XXZ XXZ ω= gα ω = hβ gα ω = gα hβ ω = ω.  P

α

Global Change of Variables.

β

α

α

β

Q

This result can now be formulated very elegantly as follows.

8.1.7 Theorem (Change of Variables Theorem). Suppose M and N are oriented n-manifolds and f : M → N is an orientation-preserving diffeomorphism. If ω ∈ Ωn (N ) has compact support, then f ∗ ω has compact support and Z Z f ∗ ω.

ω= N

Proof.

(8.1.4)

M

First, note that supp(f ∗ ω) = f −1 (supp(ω)),

which is compact. To prove equation (8.1.4), let {(Ui , ϕi )} be an atlas of positively oriented charts of M and let P = {gi } be a subordinate partition of unity. Then {(f (Ui ), ϕi ◦ f −1 )} is an atlas of positively oriented

450

8. Integration on Manifolds

charts of N and Q = {gi ◦ f −1 } is a partition of unity subordinate to the covering {f (Ui )}. By Proposition 8.1.6, Z XZ XZ ∗ ∗ f ω= gi f ω = ϕi∗ (gi f ∗ ω) M

M

i

=

XZ

Rn

ϕi∗ (f −1 )∗ (gi ◦ f −1 )ω

Rn

i

=

i

XZ

(ϕi ◦ f −1 )∗ (gi ◦ f −1 )ω

Rn

Zi =

ω.



N

This result is summarized by the following commutative diagram: f∗ Ωn (M )  Ωn (N ) f∗ @ R@ R M@ N R @ R

8.1.8 Definition. Let (M, µ) be a volume manifold. Suppose f ∈ F(M ) and f has compact support. Then R we call M f µ the integral of f with respect to µ. The reader can check that since the Riemann integral is R-linear, so is the integral just defined. Measures (Optional). The next theorem will show that the integral defined by equation (8.1.4) can be obtained in a unique way from a measure on M . (The reader unfamiliar with measure theory can find the necessary background in Royden [1968]; this result will not be essential for future sections.) The integral we have described can clearly be extended to all continuous functions with compact support. Then we have the following. 8.1.9 Theorem (Riesz Representation Theorem). Let (M, µ) be a volume manifold. Let β denote the collection of Borel sets of M , the σ-algebra generated by the open (or closed, or compact) subsets of M . Then there is a unique measure mµ on β such that for every continuous function of compact support Z

Z f dmµ =

M

f µ.

(8.1.5)

M

Proof. Existence of such a measure mµ is proved in books on measure theory, for example Royden [1968]. For uniqueness, it is enough to consider bounded open sets (by the Hahn extension theorem). Thus, let U be open in M , and let CU be its characteristic function. We construct a sequence of C ∞ functions of compact support ϕn such that ϕn ↓ CU , pointwise. Hence from the monotone convergence theorem, we conclude that Z Z Z ϕn µ = ϕn dmµ → CU dmµ = mµ (U ). Thus, mµ is unique.



8.1 The Definition of the Integral

451

The space Lp (M, µ), p ∈ R, consists of all measurable functions f such that |f |p is integrable. For p ≥ 1, the norm 1/p Z p |f | dmµ kf kp = makes Lp (M, µ) into a Banach space (functions that differ only on a set of measure zero are identified). The use of these spaces in studying objects on M itself is discussed in §8.4. The next propositions give an indication of some of the ideas. If F : M → N is a measurable mapping and mM is a measure on M , then F∗ mM is the measure on N defined by F∗ mM (A) = mM (F −1 (A)). If F is bijective, we set F ∗ (mN ) = (F −1 )∗ mN . If f : M → R is an integrable function, then f mM is the measure on M defined by Z (f mM )(A) = f dmM A

for every measurable set A in M . 8.1.10 Proposition. Suppose M and N are orientable n-manifolds with volume forms µM and µN and corresponding measures mM and mN . Let F be an orientation preserving C 1 diffeomorphism of M onto N . Then
F ∗ mN = J(µM ,µN ) F mM . (8.1.6) Proof.

Let f be any C ∞ function with compact support on M . By Theorem 8.1.7, Z Z Z Z
∗ f dmN = f µN = F (f µN ) = (f ◦ F ) J(µM ,µN ) F µM N M M ZN
= (f ◦ F ) J(µM ,µN ) F dmM . M

As in the proof of Theorem 8.1.9, this relation holds for f chosen to be the characteristic function of F (A). That is, Z
mN (F (A)) = J(µM ,µN ) F dmM .  A

Jacobians and Divergence. In preparation for the next result, we notice that on a volume manifold (M, µ), the flow Ft of any vector field X is orientation preserving for each t ∈ R (regard this as a statement on the domain of the flow, if the vector field is not complete). Indeed, since Ft is a diffeomorphism, Jµ (Ft ) is nowhere zero; since it is continuous in t and equals one at t = 0, it is positive for all t. 8.1.11 Proposition. Let M be an orientable manifold with volume form µ and corresponding measure mµ . Let X be a (possibly time-dependent) C 1 vector field on M with flow Ft . The following are equivalent (if the flow of X is not complete, the statements involving it are understood to hold on its domain): (i) divµ X = 0; (ii) Jµ Ft = 1 for all t ∈ R; (iii) Ft∗ mµ = mµ for all t ∈ R; (iv) Ft∗ µ = µ for all t ∈ R; R R (v) M f dmµ = M (f ◦ Ft )dmµ for all f ∈ L1 (M, µ) and all t ∈ R.

452

8. Integration on Manifolds

Proof. Statement (i) is equivalent to (ii) by Corollary 7.5.19. Statement (ii) is equivalent to (iii) by equation (8.1.6) and to (iv) by definition. We shall prove that (ii) is equivalent to (v). If Jµ Ft = 1 for all t ∈ R and f is continuous with compact support, then Z Z Z Z ∗ ∗ (f ◦ Ft )µ = (f ◦ Ft )(Ft µ) = Ft (f µ) = f µ. M

M

M

R

M

R

Hence, by uniqueness in Theorem 8.1.9, we have M f dmµ = M (f ◦ Ft )dmµ for all integrable f , and so (ii) implies (v). Conversely, if Z Z (f ◦ Ft )dmµ = f dmµ M

M

then taking f to be continuous with compact support, we see that Z Z Z Z (f ◦ Ft )µ = fµ = Ft∗ (f µ) = (f ◦ Ft )Ft∗ µ M

M

M

Z (f ◦ Ft )(Jµ Ft )µ.

=

M

M

Thus, for every integrable f , Z

Z (f ◦ Ft )dmµ =

M

(f ◦ Ft )(Jµ Ft )dmµ . M

Hence Jµ Ft = 1, and so (v) implies (ii).



Transport Theorem. The following result is central to continuum mechanics (see Example 8.1.13 below and §8.2 for applications). 8.1.12 Theorem (Transport Theorem). Let (M, µ) be a volume manifold and X a vector field on M with flow Ft . For f ∈ F(M × R) and letting ft (m) = f (m, t), we have d dt

Z

Z



ft µ = Ft (U )

Ft (U )

 ∂f + divµ (ft X) µ ∂t

(8.1.7)

for any open set U in M . Proof.

By the flow characterization of Lie derivatives and Proposition 7.5.17, we have   d ∗ ∂f ∗ F (ft µ) = Ft µ + Ft∗ £X (ft µ) dt t ∂t   ∂f ∗ µ + Ft∗ [(£X ft )µ + ft (divµ X)µ] = Ft ∂t    ∂f ∗ = Ft + divµ (ft X) µ . ∂t

Thus, by the change of variables formula, d dt

Z ft µ = Ft (U )

d dt Z

Z

Ft∗ (ft µ) =

U

= Ft (U )

Z U



Ft∗



  ∂f + divµ (ft X) µ ∂t

 ∂f + divµ (ft X) µ. ∂t



8.1 The Definition of the Integral

453

8.1.13 Example. Let ρ(x, t) be the density of an ideal fluid moving in a compact region of R3 with smooth boundary. One of the basic assumptions of fluid dynamics is conservation of mass: the mass of the fluid in the open set U remains unchanged during the motion described by a flow Ft . This means that Z d ρ(x, t) d3 x = 0 (8.1.8) dt Ft (U ) for all open sets U . By the transport theorem, equation (8.1.8) is equivalent to the equation of continuity ∂ρ + div(ρu) = 0, ∂t

(8.1.9)

here u represents the velocity of the fluid particles. We shall return to this example in §8.2. Recurrence.



As another application of Proposition 8.1.11, we prove the following.

8.1.14 Theorem (Poincar´e Recurrence Theorem). Let (M, µ) be a volume manifold, mµ the corresponding measure, and X a time-independent divergence-free vector field with flow Ft . Suppose A is a measurable set in M such that mµ (A) < ∞, Ft (x) exists for all t ∈ R if x ∈ A, and Ft (A) ⊂ A. Then for each measurable subset B of A and T ≥ 0, there exists S ≥ T such that B ∩ FS (B) 6= ∅. Therefore, a trajectory starting in B returns infinitely often to B. Proof. By Proposition 8.1.11, the sets B, FT (B), F2T (B), . . . all have the same finite measure. Since mµ (A) < ∞, they cannot all be disjoint, so there exist integers k > l > 0 satisfying FkT (B) ∩ FlT (B) 6= ∅. Since FkT = (FT )k (as X is time-independent), we get F(k−l)T (B) ∩ B 6= ∅.



The Poincar´e recurrence theorem is one of the forerunners of ergodic theory, a topic that will be discussed briefly in §8.4. A related result is the following. 8.1.15 Theorem (Schwarzschild Capture Theorem). Let (M, µ) be a volume manifold, X a time-independent divergence-free vector field with flow Ft , and A a measurable subset of M with finite measure. Assume that for every x ∈ A, the trajectory t 7→ Ft (x) exists for all t ∈ R. Then for almost all x ∈ A (relative to mµ ) the following are equivalent: (i) Ft (x) ∈ A for all t ≥ 0; (ii) Ft (x) ∈ A for all t ≤ 0. T Proof. Let A1 = t≥0 F Tt (A) be the set of points in A which have their future trajectory completely in A. Similarly, consider A2 = t≤0 Ft (A). By Proposition 8.1.11, for any τ ≥ 0,  µ(A1 ) = µ(F−τ (A1 )) = µ 

 \

Fτ (A)

t≥−τ

which shows, by letting τ → ∞, that ! µ(A1 ) = µ

\ t∈R

Ft (A)

= µ(A1 ∩ A2 ).

454

8. Integration on Manifolds

Reasoning similarly for A2 , we get µ(A1 ) = µ(A1 ∩ A2 ) = µ(A2 ), so that µ(A1 \(A1 ∩ A2 )) = µ(A2 \(A1 ∩ A2 )) = 0. Let S = (A1 \(A1 ∩ A2 )) ∪ (A2 \(A1 ∩ A2 )); then mµ (S) = 0 and S ⊂ A. Moreover, we have A1 S = A1 ∩ A2 = A2 S, proving the desired equivalence.  So far only integration on orientable manifolds has been discussed. A similar procedure can be carried out to define the integral of a one-density (see §8.5) on any manifold, orientable or not. The only changes needed in the foregoing definitions and propositions are to replace the Jacobians with their absolute values and to use the definition of divergence with respect to a given density as discussed in §8.5. All definitions and propositions go through with these modifications. Vector Valued Forms. If F is a finite-dimensional Pl vector space, F -valued one-forms and one-densities can also be integrated in the following way. If ω = i=1 ω i fi , where f1 , . . . , fn is an ordered basis of F , then we set  Z l Z X i ω= ω fi ∈ F. M

i=1

M

It is easy to see that this definition is independent of the chosen basis of F and that all the basic properties of the integral remain unchanged. On the other hand, the integral of vector-bundle-valued n-forms on M is not defined unless additional special structures (such as triviality of the bundle) are used. In particular, integration of vector or general tensor fields is not defined.

Exercises  8.1-1. that

Let M be an n-manifold and µ a volume form on M . If X is a vector field on M with flow Ft show d (Jµ (Ft )) = Jµ (Ft )(divµ X ◦ Ft ). dt

Hint: Compute (d/dt)Ft∗ µ using the Lie derivative formula.  8.1-2.

Prove the following generalization of the transport theorem   Z Z d ∂ωt ωt = + £X ωt , dt Ft (V ) ∂t Ft (V )

where ωt is a time–dependent k-form on M and V is a k-dimensional submanifold of M .  8.1-3. (i) Let ϕ : S 1 → S 1 be the map defined by ϕ(eiθ ) = e2iθ , where θ ∈ [0, 2π]. Let, by abuse of notion, dθ denote the standard volume of S 1 . Show that the following identity holds: Z Z ϕ∗ (dθ) = 2 dθ. S1

S1

(ii) Let ϕ : M → N be a smooth surjective map. Then ϕ called a k-fold covering map if every n ∈ N has an open neighborhood V such that ϕ−1 (V ) = U1 ∪ · · · ∪ Uk , are disjoint open sets each of which is diffeomorphic by ϕ to V . Generalize (i) in the following way. If ω ∈ Ωn (N ) is a volume form, show that Z Z ϕ∗ ω = k

M

ω. N

8.1 The Definition of the Integral

455

 8.1-4. Define the integration of Banach space valued n-forms on an n-manifold M . Show that if the Banach space is Rl , you recover the coordinate definition given at the end of this section. If E, F are Banach spaces and A ∈ L(E, F), define A∗ ∈ L(Ω(M, E), Ω(M, F)) by (A∗ α)(m) = A(α(m)). Show that Z  Z  ◦ A∗ = A ◦ M

M

on Ωn (M, E).  8.1-5. Let M and N be oriented manifolds and endow M × N with the product orientation. Let pM : M × N → M and pN : M × N → N be the projections. If α ∈ Ωdim M (M ) and β ∈ Ωdim N (N ) have compact support show that α × β := (pM ∗ α) ∧ (pN ∗ β) has compact support and is a (dim M + dim N )-form on M × N . Prove Fubini’s Theorem   Z Z Z β . α α×β = M ×N

N

M

 8.1-6 (Fiber Integral). Let ϕ : M → N be a surjective submersion, where dim M = m and dim N = n. The map ϕ is said to be orientable if there exists η ∈ Ωp (M ), where p = m − n, such that for each y ∈ N , jy ∗ η is a volume form on ϕ−1 (y), where jy : ϕ−1 (y) → M is the inclusion. An orientation of ϕ is an equivalence class of p-forms under the relation: η1 ∼ η2 iff there exists f ∈ F(M ), f > 0 such that η2 = f η1 . (i) If ϕ : M → N is a vector bundle, show that orientability of ϕ is equivalent to orientability of the vector bundle as defined in Exercise 7.5-13. (ii) If ϕ is oriented by η and N by ω, show that ϕ∗ ω ∧ η is a volume form on M . The orientation on M defined by this volume is called the local product orientation of M (compare with Exercise 7.5-13(vi)). (iii) Let Ωkϕ (M ) := { α ∈ Ωk (M ) | ϕ−1 (K) ∩ supp(α) is compact, for any compact set K ⊂ N }, the fiber-compactly supported k-forms on M . Show that Ωkϕ (M ) is an F(M )-submodule of Ωk (M ), and is invariant under the interior product, exterior differential, and Lie derivative. −1 (iv) If α ∈ Ωk+p (y), with values in Ty∗ N ∧ · · · ∧ Ty∗ N (k ϕ (M ), k ≥ 0 and y ∈ N , define a p-form αy on ϕ times) by

[αy (x)(u1 , . . . , up )](Tx ϕ(v1 ), . . . , Tx ϕ(vk )) = α(x)(v1 , . . . , vk , u1 , . . . , up ), where ϕ(x) = y, x ∈ M , v1 , . . . , vk ∈ Tx M, and u1 , . . . , up ∈ ker(Tx ϕ) = Tx (ϕ−1 (y)). Assume ϕ is oriented. Define the fiber integral Z k : Ωk+p ϕ (M ) → Ω (N ) fib

by Z fib

 Z α (x) =

αy ,

if ϕ(x) = y;

ϕ−1 (y)

the right-hand side is understood as the integral of a vector-valued p-form on the oriented p-manifold ϕ−1 (y).

456

8. Integration on Manifolds

R (a) Prove that fib α is a smooth k-form on N . Hint: Use charts in which ϕ is a projection and apply the theorem of smoothness of the integral with respect to parameters. R (b) Show that if ϕ : M → N is a locally trivial fiber bundle with N paracompact, fib is surjective. ∗ k+l+p (v) Let β ∈ Ωl (N ) have compact support and let α ∈ Ωk+p (M ) and ϕ (M ). Show that ϕ β ∧ α ∈ Ωϕ that Z Z (ϕ∗ β ∧ α) = β ∧ α. fib

fib

Hint: Let E = Ty∗ N ∧ · · · ∧ Ty∗ N (k times) and let F be the wedge product l + k times. Define ∗ A ∈ L(E, F ) by A(γ) R = β(y) ∧ γ and show that (ϕ β ∧ α)y = A∗ (αy ) using the notation of Exercise 8.1-4. Then apply fib to this identity and use Exercise 8.1-4. (vi) Assume N is paracompact and oriented, ϕ is oriented, and endow M with the local product orientation. Prove the following iterated integration (Fubini-type) formula Z Z Z = ◦ M

N

fib

by following the three steps below. Step 1: Using a partition of unity, reduce to the case M = N × P where ϕ : M → N is the projection and M, N, P are Euclidean spaces. Step 2: Use (v) and Exercise 8.1-5 to show that for β ∈ Ωn (N ) and γ ∈ Ωp (P ) with compact support, Z Z Z (β × γ) = (β × γ) N

fib

M

Step 3: Since M, N, and P are ranges of coordinate patches, show that any ω ∈ Ωm (M ) with compact support is of the form β × γ. 0 (vii) Let ϕ : M → N and ϕ0 : M 0 → N 0 be oriented surjective submersions R and let f : M R 0 → M , and R0 f0 : N → N 0 be smooth maps satisfying f0 ◦ ϕ = ϕ0 ◦ f. Show that fib ◦f ∗ = f0∗ ◦ fib , where fib denotes the fiber integral of ϕ0 .

(viii) Let ϕ : M → N be an oriented surjective submersion and assume X ∈ X(M ) and Y ∈ X(N ) are ϕ-related. Prove that Z Z Z Z Z Z ◦ iX = iY ◦ , ◦d = d ◦ , ◦ £X = £Y ◦ . fib

fib

fib

fib

fib

fib

(For more information on the fiber integral see Bourbaki [1971] and Greub, Halpern and Vanstone.)  8.1-7. Let ϕ : M → N be a smooth orientation preserving map, where M and N are volume manifolds of dimension m and n respectively. For α ∈ Ωk (M ) with compact support, define the linear functional ϕ∗ α : Ωm−k → R by Z ϕ∗ β ∧ α

(ϕ∗ α)(β) = M

for all β ∈ Ωm−k (N ); that is, ϕ∗ α is a distributional k-form on N . If m < k, set ϕ∗ α = 0. If there is a γ ∈ Ωn−m+k (N ) satisfying Z (ϕ∗ α)(β) = β ∧ γ, M

identify ϕ∗ α with γ and say ϕ∗ α is of form-type. Prove the following statements.

8.1 The Definition of the Integral

457

(i) If ϕ is a diffeomorphism, then ϕ∗ α is the usual push-forward. (ii) If α is a volume form, this definition corresponds to that for the push-forward of measures. R (iii) If ϕ is an oriented surjective submersion, show that ϕ∗ α = fib α, as defined in Exercise 8.1-6(iv). Hint: Prove the identity Z

ϕ∗ β ∧ α =

Z 

M



Z β∧

N

α fib

using Exercise 8.1-6(v) and (vi).  8.1-8.

Let (M, µ) be a paracompact n-dimensional volume manifold.

(i) If (N, ν) is another paracompact n-dimensional R R volume manifold and f : M → N is an orientation reversing diffeomorphism, show that N ω = − M f ∗ ω for any ω ∈ Ωn (N ) with compact support. Hint: Use the proof of Theorem 8.1.12. (ii) If η ∈ Ωn (M ) has compact support and −M denotes the manifold M endowed with the orientation [−µ], show that Z Z η=− η. −M

M

Hint: If A = {(Ui , ϕi )} is an oriented atlas for (M, [µ]), then −A = {(Ui , ϕi ◦ ψi )},

ψi (x1 , . . . , xn ) = (−x1 , x2 , . . . , xn )

is an oriented atlas for (M, [−µ]).  8.1-9.

Let ωn be the standard volume form on S n . Show that Z 2m+1 π m , if n = 2m, m ≥ 1 ωn = (2m − 1)!! Sn

and Z ωn = Sn

2π m+1 , m!

if n = 2m + 1, m ≥ 0

using the following steps. (i) Let M ⊂ Rn+1 be the annulus 

x ∈ Rn+1 | 0 < a < kxk < b < ∞



and let f : ]a, b[ × S n → A be the diffeomorphism f (t, s) = ts. Use Exercise 7.5-19(ii) to show that for x ∈ Rn+1 ,   2 2 f ∗ e−kxk Ωn+1 = tn e−t (dt × ωn ) where Ωn+1 = e1 ∧ · · · ∧ en+1 for {e1 , . . . , en+1 } the standard basis of Rn+1 , and where dt × ωn denotes the product volume form on ]a, b[ × S n . (ii) Deduce the equality Z

−kxk2

e Rn+1

Z Ωn+1 =

b

n −t2

t e a

Z dt

ωn . Sn

458

8. Integration on Manifolds

(iii) Let a ↓ 0 and b ↑ ∞ to deduce the equality Z

∞

2

tn d−t dt

Z

∞

Sn

2

e−u du =

√

+∞

Z

∞

2

t2m e−t dt =

0

Z

∞

2

t2m+1 e−t =

0

to deduce the required formula for

8.2

R Sn

n+1 .

−∞

π,

−∞

and

2

e−u du

ωn =

0

Prove that

Z

Z

√ (2m − 1)!! π , 2m+1

m! , 2

ωn .

Stokes’ Theorem

Stokes’ theorem states that if α is an (n − 1)-form on an orientable n-manifold M , then the integral of dα over M equals the integral of α over ∂M , the boundary of M . As we shall see in the next section, the classical theorems of Gauss, Green, and Stokes are special cases of this result. Before stating Stokes’ theorem formally, we need to discuss manifolds with boundary and their orientations. 8.2.1 Definition.

Let E be a Banach space and λ ∈ E∗ . Let Eλ = { x ∈ E | λ(x) ≥ 0 },

called a half-space of E, and let U ⊂ Eλ be an open set (in the topology induced on Eλ from E). Call Int U = U ∩ { x ∈ E | λ(x) > 0 } the interior of U and ∂U = U ∩ ker λ the boundary of U . If E = Rn and λ is the projection on the jth factor, then Eλ is denoted by Rnj and is called positive jth half-space. Rnn is also denoted by Rn+ . We have U = Int U ∪ ∂U , Int U is open in U , ∂U is closed in U (not in E), and ∂U ∩ Int U = ∅. The situation is shown in Figure 8.2.1. Note that ∂U is not the topological boundary of U in E, but it is the topological boundary of U intersected with that of Eλ . This inconsistent use of the notation ∂U is temporary.

Figure 8.2.1. Open sets in a half-space

A manifold with boundary will be obtained by piecing together sets of the type shown in the figure. To carry this out, we need a notion of local smoothness to be used for overlap maps of charts. 8.2.2 Definition. Let E and F be Banach spaces, λ ∈ E∗ , µ ∈ F∗ , U be an open set in Eλ , and V be an open set in Fµ . A map f : U → V is called smooth if for each point x ∈ U there are open neighborhoods

8.2 Stokes’ Theorem

459

U1 of x in E and V1 of f (x) in F and a smooth map f1 : U1 → V1 such that f |U ∩ U1 = f1 |U ∩ U1 . We define Df (x) = Df1 (x). The map f is a diffeomorphism if there is a smooth map g : V → U which is an inverse of f . (In this case Df (x) is an isomorphism of E with F.) We must prove that this definition of Df is independent of the choice of f1 , that is, we have to show that if ϕ : W → E is a smooth map with W open in E such that ϕ|(W ∩ Eλ ) = 0, then Dϕ(x) = 0 for all x ∈ W ∩ Eλ . If x ∈ Int(W ∩ Eλ ), this fact is obvious. If x ∈ ∂(W ∩ Eλ ), choose a sequence xn ∈ Int(W ∩ Eλ ) such that xn → x; but then 0 = Dϕ(xn ) → Dϕ(x) and hence Dϕ(x) = 0, which proves our claim. 8.2.3 Lemma. Let U ⊂ Eλ be open, ϕ : U → Fµ be a smooth map, and assume that for some x0 ∈ Int U , ϕ(x0 ) ∈ ∂Fµ . Then Dϕ(x0 )(E) ⊂ ∂Fµ = ker µ. Proof. The quotient F/ ker µ is isomorphic to R, so that fixing f with µ(f ) > 0, the element [f ] ∈ F/ ker µ forms a basis. Therefore [f ] determines the isomorphism Tf : F/ ker µ → R given by Tf ([y]) = t, where t ∈ R is the unique number for which t[f ] = [y]. This isomorphism in turn defines the isomorphism Sf : ker µ ⊕ R → F given by Sf (y, t) = y + tf which induces diffeomorphisms (in the sense of Definition 8.2.2) of ker µ × [0, ∞[ with Fµ and of ker µ × ]−∞, 0] with { y ∈ F | µ(y) ≤ 0 }. Denote by p : F → R the linear map given by Sf−1 followed by the projection ker µ ⊕ R → R, so that y ∈ Fµ (respectively, ker µ, { y ∈ F | µ(y) ≤ 0 }) if and only if p(y) ≥ 0 (respectively, = 0, ≤ 0). Notice that the relation ϕ(x0 + tx) = ϕ(x0 ) + Dϕ(x0 ) · tx + o(tx), where limt→0 o(tx)/t = 0, together with the hypothesis (p ◦ ϕ)(x) ≥ 0 for all x ∈ U, implies that 0 ≤ (p ◦ ϕ)(x0 + tx) = 0 + (p ◦ Dϕ)(x0 ) · tx + p(o(tx)), whence for t > 0  0 ≤ (p ◦ Dϕ)(x0 ) · x + p

o(tx) t

 .

Letting t → 0, we get (p ◦ Dϕ)(x0 ) · x ≥ 0 for all x ∈ E. Similarly, for t < 0 and letting t → 0, we get (p ◦ Dϕ)(x0 ) · x ≤ 0 for all x ∈ E. The conclusion is (Dϕ)(x0 )(E) ⊂ ker µ.



Intuitively, this says that if ϕ preserves the condition λ(x) ≥ 0 and maps an interior point to the boundary, then the derivative must be zero in the normal direction. The reader may also wish to prove Lemma 8.2.3 from the implicit mapping theorem. Now we carry this idea one step further. 8.2.4 Lemma. Let U be open in Eλ , V be open in Fµ , and f : U → V be a diffeomorphism. Then f restricts to diffeomorphisms Int f : Int U → Int V and ∂f : ∂U → ∂V. Proof. Assume first that ∂U = ∅, that is, that U ∩ ker λ = ∅. We shall show that ∂V = ∅ and hence we take Int f = f. If ∂V 6= ∅, there exists x ∈ U such that f (x) ∈ ∂V and hence by definition of smoothness there are open neighborhoods U1 ⊂ U and V1 ⊂ F, such that x ∈ U1 and f (x) ∈ V1 , and smooth maps f1 : U1 → V1 , g1 : V1 → U1 such that f |U1 = f1 ,

g1 |V ∩ V1 = f −1 |V ∩ V1 .

460

8. Integration on Manifolds

Let xn ∈ U1 , xn → x, yn ∈ V1 \∂V , and yn = f (xn ). We have Df (x) ◦ Dg1 (f (x)) = =

lim (Df (g1 (yn )) ◦ Dg1 (yn ))

yn →f (x)

lim

yn →f (x)

D(f ◦ g1 )(yn ) = IdF

and similarly Dg1 (f (x)) ◦ Df (x) = IdE so that Df (x)−1 exists and equals Dg1 (f (x)). But by Lemma 8.2.3, Df (x)(E) ⊂ ker µ, which is impossible, Df (x) being an isomorphism. Assume that ∂U 6= ∅. If we assume ∂V = ∅, then, working with f −1 instead of f , the above argument leads to a contradiction. Hence ∂V 6= ∅. Let x ∈ Int U so that x has a neighborhood U1 ⊂ U , U1 ∩ ∂U = ∅, and hence ∂U1 = ∅. Thus, by the preceding argument, ∂f (U1 ) = ∅, and f (U1 ) is open in V \∂V . This shows that f (Int U ) ⊂ Int V . Similarly, working with f −1 , we conclude that f (Int U ) ⊃ Int V and hence f : Int U → Int V is a diffeomorphism. But then f (∂U ) = ∂V and f |∂U : ∂U → ∂V is a diffeomorphism as well.  Now we are ready to define a manifold with boundary. 8.2.5 Definition. A manifold with boundary is a set M together with an atlas of charts with boundary on M ; charts with boundary are pairs (U, ϕ) where U ⊂ M and ϕ(U ) ⊂ Eλ for some λ ∈ E ∗ and an atlas on M is a family of charts with boundary satisfying MA1 and MA2 of Definition 3.1.1, with smoothness of overlap maps ϕji understood in the sense of Definition 8.2.2. See Figure 8.2.2. If E = Rn , M is called an n-manifold with boundary . Define [ [ Int M = ϕ−1 (Int(ϕ(U ))) and ∂M = ϕ−1 (∂(ϕ(U ))) U

U

called, respectively, the interior and boundary of M . The definition of Int M and ∂M makes sense in view of Lemma 8.2.4. Note that 1. Int M is a manifold (with atlas obtained from (U, ϕ) by replacing ϕ(U ) ⊂ Eλ by the set Int ϕ(U ) ⊂ E); 2. ∂M is a manifold (with atlas obtained from (U, ϕ) by replacing ϕ(U ) ⊂ Eλ by ∂ϕ(U ) ⊂ ∂E = ker λ); 3. ∂M is the topological boundary of Int M in M (although Int M is not the topological interior of M ). Summarizing, we have proved the following. 8.2.6 Proposition. If M is a manifold with boundary, then its interior Int M and its boundary ∂M are smooth manifolds without boundary. Moreover, if f : M → N is a diffeomorphism, N being another manifold with boundary, then f induces, by restriction, two diffeomorphisms Int f : Int M → Int N

and

∂f : ∂M → ∂N.

If n = dim M , then dim(Int M ) = n and dim(∂M ) = n − 1. To integrate a differential n-form over an n-manifold M , M must be oriented. If Int M is oriented, we want to choose an orientation on ∂M compatible with it. In the classical Stokes theorem for surfaces, it is crucial that the boundary curve be oriented, as in Figure 8.2.3. The tangent bundle to a manifold with boundary is defined in the same way as for manifolds without boundary. Recall that any tangent vector in Tx M has the form [dc(t)/dt]|t=τ , where c : [a, b] → M is a C 1 curve, a < b, and τ ∈ [a, b]. If x ∈ ∂M , we consider curves c : [a, b] → M such that c(b) = x. If ϕ : U → U 0 ⊂ Eλ is a chart at m, then [d(ϕ ◦ c)(t)/dt]|t=b in general points out of U 0 , as in Figure 8.2.4.

8.2 Stokes’ Theorem

461

xn

Figure 8.2.2. Boundary charts

Figure 8.2.3. Orientation for surfaces

Therefore, Tx M is isomorphic to the model space E of M even if x ∈ ∂M (see Figure 8.2.4). It is because of this result that tangent vectors are derivatives of C 1 -curves defined on closed intervals. Had we defined tangent vectors as derivatives of C 1 -curves defined on open intervals, Ty M for y ∈ ∂M would be isomorphic to ker λ and not to E. In Figure 8.2.4, E = Rn , λ is the projection onto the n-th factor, and c(t) is defined on a closed interval whereas the C 1 -curve d(t) is defined on an open interval. M

x

y

Figure 8.2.4. Tangent spaces at the boundary.

Having defined the tangent bundle, all of our previous constructions including tensor fields and exterior forms as well as operations on them such as the Lie derivative, interior product, and exterior derivative carry over directly to manifolds with boundary. One word of caution though: the fundamental relation between Lie derivatives and flows still holds if one is careful to take into account that a vector field on M has integral

462

8. Integration on Manifolds

curves which could run into the boundary in finite time and with finite velocity. (If the vector field is tangent to ∂M , this will not happen.)

Figure 8.2.5. Oriented boundary charts

Next, we turn to the problem of orientation. As for manifolds without boundary a volume form on an n-manifold with boundary M is a nowhere vanishing n-form on M . Fix an orientation on Rn . Then a chart (U, ϕ) is called positively oriented if Tu ϕ : Tu M → Rn is orientation preserving for all u ∈ U . If M is paracompact this latter condition is equivalent to orientability of M (the proof is as in Proposition 7.5.2). Therefore, for paracompact manifolds, an orientation on M is just a smooth choice of orientations of all the tangent spaces, “smooth” meaning that for all the charts of a certain atlas, called the oriented charts, the n n maps D(ϕj ◦ ϕ−1 i )(x) : R → R are orientation preserving. The reader may wonder why for finite dimensional manifolds we did not choose a “standard” half-space, like xn ≥ 0 to define the charts at the boundary. Had we done that, the very definition of an oriented chart would be in jeopardy. For example, consider M = [0, 1] and agree that all charts must have range in R+ = { x ∈ R | x ≥ 0 }. Then an example of an orientation reversing chart at x = 1 is ϕ(x) = 1 − x; in fact, every chart at x = 1 would be orientation reversing. However, if we admit any half-space of R, so charts can be also in R− = {x ∈ R | x ≤ 0}, then a positively oriented chart at 1 is ϕ(x) = x − 1. See Figure 8.2.6.

x

x 0

0

1

x

x

0 Figure 8.2.6. Boundary charts for [0, 1]

Once oriented charts and atlases are defined, the theory of integration for oriented paracompact manifolds with boundary proceeds as in §8.1. Finally we define the boundary orientation of ∂M . At every x ∈ ∂M , the linear space Tx (∂M ) has codimension one in Tx M so that there are (in a chart on M intersecting ∂M ) exactly two kinds of vectors not in ker λ: those for which their representatives v satisfy λ(v) > 0 or λ(v) < 0, that is, the inward and outward pointing vectors. By Lemma 8.2.4, a change of chart does not affect the property of a vector being outward or inward (see Figure 8.2.4). If dim M = n, these considerations enable us to define the induced orientation of ∂M in the following way.

8.2 Stokes’ Theorem

463

8.2.7 Definition. Let M be an oriented n-manifold with boundary, x ∈ ∂M and ϕ : U → Rnλ a positively oriented chart, where λ ∈ (Rn )∗ . A basis {v1 , . . . , vn−1 } of Tx (∂M ) is called positively oriented if  (Tx ϕ)−1 (n), v1 , . . . , vn−1 is positively oriented in the orientation of M , where n is any outward pointing vector to Rnλ at ϕ(x). For example, we could choose for n the outward pointing vector to Rnλ and perpendicular to ker λ. If λ : Rn → R is the projection on the n-th factor, then (Tx ϕ)−1 (n) = −∂/∂xn and the situation is illustrated in Figure 8.2.5. 8.2.8 Theorem (Stokes’ Theorem). Let M be an oriented smooth paracompact n-manifold with boundary and α ∈ Ωn−1 (M ) have compact support. Let i : ∂M → M be the inclusion map so that i∗ α ∈ Ωn−1 (∂M ). Then Z Z i∗ α = dα (7.2.1a) ∂M

M

or for short, Z

Z α=

∂M

dα

(7.2.1b)

M

If ∂M = ∅, the left hand side of equation (7.2.1a) or (7.2.1b) is set equal to zero. Proof. Since integration was constructed with partitions of unity subordinate to an atlas and both sides of the equation to be proved are linear in α, we may assume without loss of generality that α is a form on U ⊂ Rn+ with compact support. Write α=

n X

(−1)i−1 αi dx1 ∧ · · · ∧ (dxi )ˆ ∧ · · · ∧ dxn ,

(8.2.2)

i=1

where ˆ above a term means that it is deleted. Then dα =

n X ∂αi i=1

and thus

∂xi

dx1 ∧ · · · ∧ dxn ,

(8.2.3)

n Z X

∂αi 1 dx · · · dxn . (8.2.4) i n ∂x U R i=1 R There are two cases: ∂U = ∅ and ∂U 6= ∅. If ∂U = ∅, we have ∂U α = 0. The integration of the ith term in the sum occurring in equation (8.2.4) is  Z Z ∂αi i dx dx1 · · · (dxi )ˆ· · · dxn (no sum) (8.2.5) i Rn−1 R ∂x Z

dα =

R +∞ and −∞ (∂αi /∂xi )dx = 0 since αi has a compact support. Thus, the expression in equation (8.2.4) is zero as desired. If ∂U 6= ∅, then we can do the same trick for each term except the last, which is, by the fundamental theorem of calculus, Z ∞  Z ∂αn n dx dx1 · · · dxn−1 ∂xn Rn−1 0 Z =− αn (x1 , . . . , xn−1 , 0)dx1 · · · dxn−1 . (8.2.6) Rn−1

464

8. Integration on Manifolds

since αn has compact support. Thus, Z Z dα = −

αn (x1 , . . . , xn−1 , 0)dx1 · · · dxn−1

(8.2.7)

Rn−1

U

On the other hand, Z

Z

Z

α=

α= ∂Rn +

∂U

(−1)n−1 αn (x1 , . . . , xn−1 , 0)dx1 ∧ · · · ∧ dxn−1 .

(8.2.8)

∂Rn +

But Rn−1 = ∂Rn+ and the usual orientation on Rn−1 is not the boundary orientation. The outward unit normal is −en = (0, . . . , 0, −1) and hence the boundary orientation has the sign of the ordered basis {−en , e1 , . . . , en−1 }, which is (−1)n . Thus equation (8.2.8) becomes Z Z α= (−1)n−1 αn (x1 , . . . , xn−1 , 0)dx1 ∧ · · · ∧ dxn−1 ∂U

∂Rn + 2n−1

Z

= (−1)

αn (x1 , . . . , xn−1 , 0)dx1 · · · dxn−1 .

(8.2.9)

Rn−1

Since (−1)2n−1 = −1, combining equations (8.2.7) and (8.2.9), we get the desired result.



This basic theorem reduces to the usual theorems of Green, Stokes, and Gauss in R2 and R3 , as we shall see in the next section. For forms with less smoothness or without compact support, the best results are somewhat subtle. See Gaffney [1954], Morrey [1966], Yau [1976], Karp [1981] and the remarks at the end of Supplement 8.2B. Next we draw some important consequences from Stokes’ theorem. 8.2.9 Theorem (Gauss’ Theorem). Let M be an oriented paracompact n-manifold with boundary and X a vector field on M with compact support. Let µ be a volume form on M . Then Z Z (div X)µ = iX µ. (8.2.10) M

Proof.

∂M

Recall that (div X)µ = £X µ = diX µ + iX dµ = diX µ.

The result is thus a consequence of Stokes’ theorem.



If M carries a Riemannian metric, there is a unique outward–pointing unit normal n∂M along ∂M , and M and ∂M carry corresponding uniquely determined volume forms µM and µ∂M . Then Gauss’ theorem reads as follows. 8.2.10 Corollary. Z

Z hhX, n∂M ii dµ∂M ,

(div X) dµM = M

∂M

where hhX, n∂M ii is the inner product of X and n∂M is the outward unit normal. Proof. Let µ∂M denote the volume element on ∂M induced by the Riemannian volume element µM ∈ Ωn (M ); that is, for any positively oriented basis v1 , . . . , vn−1 ∈ Tx (∂M ), and charts chosen so that n∂M = −∂/∂xn at the point x,   ∂ µ∂M (x)(v1 , . . . , vn−1 ) = µM (x) − n , v1 , . . . , vn−1 . ∂x

8.2 Stokes’ Theorem

465

Since 

∂ (iX µM )(x)(v1 , . . . , vn−1 ) = µM (x) X (x)vi + X (x) n , v1 , . . . , vn−1 ∂x = X n (x)µ∂M (x)(v1 , . . . , vn−1 ) i

n

and X n = −hhX, n∂M ii, the corollary follows by Gauss’ theorem.





8.2.11 Corollary. If X is divergence-free on a compact boundaryless manifold with a volume element µ, then X as an operator is skew-symmetric; that is, for f and g ∈ F(M ), Z Z X[f ]gµ = − f X[g]µ. M

Proof.

M

Since X is divergence free, £X (hµ) = (£X h)µ for any h ∈ F(M ). Thus, X[f ]gµ + f X[g]µ = £X (f g)µ = £X (f gµ).

Integration and the use of Stokes’ theorem gives the result.



If M is compact without boundary X ∈ X(M ), α ∈ Ωk (M ), and β ∈ Ωn−k (M ), then Z Z £X α ∧ β = − α ∧ £X β.

8.2.12 Corollary.

M

M

Proof. Since α ∧ β ∈ Ωn (M ), the formula follows by integrating both sides of the relation diX (α ∧ β) = £X (α ∧ β) = £X α ∧ β + α ∧ £X β and using Stokes’ theorem.  8.2.13 Corollary. If M is a compact orientable, boundaryless n-dimensional pseudo-Riemannian manifold with a metric g of index Ind(g), then d and δ are adjoints, that is, Z Z Z Z hdα, βiµ = dα ∧ ∗β = α ∧ ∗δβ = hα, δβiµ M

M

M

M

for α ∈ Ωk (M ) and β ∈ Ωk+1 (M ). Proof.

Recall from Definition 7.5.21 that δβ = (−1)nk+1+Ind(g) ∗ d ∗ β,

so that dα ∧ ∗β − α ∧ ∗δβ = dα ∧ ∗β + (−1)nk+Ind(g) α ∧ ∗ ∗ d ∗ β = dα ∧ ∗β + (−1)nk+Ind(g)+k(n−k)+Ind(g) α ∧ d∗ β = dα ∧ ∗β + (−1)k α ∧ d ∗ β = d(α ∧ ∗β) since k 2 + k is an even number for any integer k. Integrating both sides of the equation and using Stokes’ theorem gives the result.  The same identity ZZ

Z hdα, βiµ =

hα, δβiµ M

holds for noncompact manifolds, possibly with boundary, provided either α or β has compact support in Int(M ).

466

8. Integration on Manifolds

Supplement 8.2A Stokes’ Theorem for Nonorientable Manifolds Let M be a nonorientable paracompact n-manifold with a smooth boundary ∂M and inclusion map i : ∂M → M . We would like to give meaning to the formula Z Z dρ = i∗ ρ. M

∂M

in Stokes’ theorem. Clearly, both sides makes sense if dρ and i∗ (ρ) are defined in such a way that they are densities on M and ∂M , respectively. Here d should be some operator Vn−1 analogous to the exterior differential, and ρ should be a section of some bundle over M analogous to (M ). Denote the as yet unknown bundle Vk Vk k analogous to (M ) by t (M ) and its space of sections Ω (M ). Then we desire an operator d : Ωkτ (M ) → τ Vn V k+1 Ωτ (M ), k = 0, . . . , n, and desire τ (M ) to be isomorphic to | |(M ). Vk Vn To guess what τ (M ) might be, let us first discuss τ (M ). The key difference between an n-form ω and a density ρ is their transformation property under a linear map A : Tm M → Tm M as follows: ω(m)(A(v1 ), . . . , A(vn )) = (det A) ω(m)(v1 , . . . , vn ) ρ(m)(A(v1 ), . . . , A(vn )) = | det A| ρ(m)(v1 , . . . , vn ) for m ∈ M and v1 , . . . , vn ∈ Tm M . If v1 , . . . , vn is a basis, then det(A) > 0 if A preserves the orientation given by v1 , . . . , vn and det(A) < 0 if A reverses this orientation. Thus ρ can be thought of as an object behaving like an n-form at every m ∈ M once an orientation of Tm M is given; that is, ρ should be thought of as an n-form with values in some line bundle (a bundle with one-dimensional fibers) associated with the Vk concept of orientation. This definition would then generalize to any k; τ (M ) will be line-bundle-valued k-forms on M . We shall now construct this line bundle. At every point of M there are two orientations. Using them, we construct the oriented double covering f → M (see Proposition 7.5.7). Since M f is not a line bundle, some other construction is in order. At every M m ∈ M , a line is desired such that the positive half-line should correspond to one orientation of Tm M and the negative half-line to the other. The fact that must be taken into account is that multiplication by a negative number switches these two half-lines. To incorporate this idea, identify (m, [µ], a) with (m, [−µ], −a) where m ∈ M , a ∈ R, and [µ] is an orientation of Tm M . Thus, define the orientation line bundle σ(M ) = { (m, [µ], a) | m ∈ M, a ∈ R, and [µ] is an orientation of Tm M ) }/∼ where ∼ is the equivalence relation (m, [µ], a) ∼ (m, [−µ], −a). Denote by hm, [µ], ai the elements of σ(M ). It can be checked that the map π : σ(M ) → M defined by π(hm, [µ], ai) = m is a line bundle with bundle charts given by ψ : π −1 (U ) → ϕ(U ) × R,

ψ(hm, [µ], ai) = (ϕ(m), a),

where ϕ : U → Rn is a chart for M at m, and  = +1 if Tm ϕ : (Tm M, [µ]) → (Rn , [ω]) is orientation preserving and −1 if it is orientation reversing, [ω] being a fixed orientation of Rn . The change of chart map of the line bundle σ(M ) is given by (x, a) ∈ U 0 × R ⊂ Rn × R
0 7→ (ϕj ◦ ϕi )−1 (x), sign(det D(ϕj ◦ ϕ−1 i )(x)) ∈ U × R. If M is paracompact, then σ(M ) is an orientable vector bundle (see Exercise 7.5-14). If in addition M is also connected, then M is orientable if and only if σ(M ) is trivial line bundle; the proof is similar to that of Proposition 7.5.7.

8.2 Stokes’ Theorem

467

8.2.14 Definition. A twisted k-form on M is a σ(M )-valued k-form on M . The bundle of twisted Vk Vk k-forms is denoted by τ (M ) and sections of this bundle are denoted Ωkτ (M ) or Γ∞ ( τ (M )). Locally, a section ρ ∈ Ωkτ (M ) can be written as ρ = αξ where α ∈ Ωk (U ) and ξ is an orientation of U regarded as a locally constant section of σ(M ) over U . The operators d : Ωkτ (M ) → Ωk+1 (M ) and τ iX : Ωkτ (M ) → Ωk−1 (M ), where X ∈ X(M ), τ are defined to be the unique operators such that if ρ = αξ in the neighborhood U , then dρ = (dα)ξ and Vk Vk iX ρ = (iX α)ξ. One has £X = iX ◦ d +V d ◦ iX . NoteVthat if M is orientable, τ (MV) coincides with (M ). n Next we show that the line bundles | (M )| and τ (M ) are isomorphic. If λ ∈ | (M )|m and v1 , . . . , vn ∈ Tm M , define ϕ(λ) : Tm M × · · · × Tm M → σ(M )m by setting Φ(λ)(v1 , . . . , vn ) = hm, [σ(v1 , . . . , vn )], λ(v1 , . . . , vn )i, if {v1 , . . . , vn } is a basis of Tm M, and setting it equal to 0, if {v1 , . . . , vn } are linearly dependent, where [σ(v1 , . . . , vn )] denotes the orientation of Tm M given by the ordered basis {v1 , . . . , vn }. Φ(λ) is skew symmetric and homogeneous with respect to scalar multiplication since if {v1 , . . . , vn } is a basis and a ∈ R, we have Φ(λ)(v2 , v1 , v3 , . . . , vn ) = hm, [σ(v2 , v1 , v3 , . . . , vn )], λ(v2 , v1 , v3 , . . . , vn )i = hm, [−σ(v1 , v2 , . . . , vn )], λ(v1 , . . . , vn )i = hm, [σ(v1 , . . . , vn )], −λ(v1 , . . . , vn )i = −Φ(λ)(v1 , . . . , vn ) and Φ(λ)(av1 , v2 , . . . , vn ) = hm, [σ(av1 , v2 , . . . , vn )], λ(av1 , . . . , vn )i = hm, [(sign a)σ(v1 , . . . , vn )], |a|λ(v1 , . . . , vn )i = hm, [σ(v1 , . . . , vn )], aλ(v1 , . . . , vn )i = aΦ(λ)(v1 , . . . , vn ). The proof of additivity is more complicated. Let v1 , v10 , v2 , . . . , vn ∈ Tm M . If both {v1 , . . . , vn } and {v10 , v2 , . . . , vn } are linearly dependent, then so are {v1 +v10 , v2 , . . . , vn } and the additivity property of Φ(λ) is trivially verified. So assume that {v1 , . . . , vn } is a basis of Tm M and write v10 = a1 v1 + · · · + an vn . Therefore λ(v10 , v2 , . . . , vn ) = |a1 |λ(v1 , . . . , vn ), and λ(v1 + v10 , v2 , . . . , vn ) = |1 + a1 |λ(v1 , . . . , vn ). Moreover, if (i) a1 > 0, then [σ(v1 , . . . , vn )] = [σ(v10 , v2 , . . . , vn )] = [σ(v1 + v10 , v2 , . . . , vn )]; (ii) a1 = 0, then [σ(v1 , . . . , vn )] = [σ(v1 + v10 , v2 , . . . , vn )] and Φ(λ)(v10 , v2 , . . . , vn ) = 0;

468

8. Integration on Manifolds

(iii) −1 < a1 < 0, then [σ(v1 , . . . , vn )] = [−σ(v10 , v2 , . . . , vn )] = [σ(v1 + v10 , v2 , . . . , vn )]; (iv) a1 = −1, then [σ(v1 , . . . , vn )] = [−σ(v10 , v2 , . . . , vn )] and Φ(λ)(v1 + v10 , v2 , . . . , vn ) = 0; (v) a1 < −1, then [σ(v1 , . . . , vn )] = [−σ(v10 , v2 , . . . , vn )] = [−σ(v1 + v10 , v2 , . . . , vn )]. Additivity is now checked in all five cases separately. For example, in case (iii) we have Φ(λ)(v1 + v10 , v2 , . . . , vn ) = hm, [σ(v1 + v10 , v2 , . . . , vn )], λ(v1 + v10 , . . . , vn )i = hm, [σ(v1 , . . . , vn )], (1 + a1 )λ(v1 , . . . , vn )i = hm, [σ(v1 , . . . , vn )], λ(v1 , . . . , vn )i +hm, [−σ(v10 , v2 , . . . , vn )], −|a1 |λ(v1 , . . . , vn )] = Φ(λ)(v1 , . . . , vn ) + Φ(λ)(v10 , v2 , . . . , vn ). Vn Thus V Φ has values Vn in τ (M ). The map Φ is clearly linear and injective and thusn is an isomorphism of | (M )| with τ (M ). Denote also by Φ the induced isomorphism of |Ω(M )| with Ωτ (M ). The integral of ρ ∈ Ωnτ (M ) is defined to be the integral of the density Φ−1 (ρ) over M . In local coordinates the expression for Φ is Φ(a|dx1 ∧ · · · ∧ dxn |) = (a dx1 ∧ · · · ∧ dxn )ξ0n , where ξ0n is the basis element of the space sections of σ(U ) given by ξ0n (u)(v1 , . . . , vn ) = hu, [σ(v1 , . . . , vn )], sign(det(vji ))i, where (vji ) are the components of the vector vj in the coordinates (x1 , . . . , xn ) of U . Therefore aξ |dx1 ∧ · · · ∧ dxn | ξ0n

Φ−1 ((a dx1 ∧ · · · ∧ dxn )ξ) = and

Z

1

n

(a dx ∧ · · · ∧ dx U

)bξ0n

Z =

ab|dx1 ∧ · · · ∧ dxn |

U

for any smooth functions a, b : U → R. Finally, for the formulation of Stokes’ Theorem, if i : ∂M → M is the inclusion and ρ ∈ Ωn−1 (M ), the τ induced twisted (n − 1)-form i∗ ρ on ∂M is defined by setting (i∗ ρ)(m)(v1 , . . . , vn−1 ) = hm, [sign[µn ]σ(−∂/∂xn , v1 , . . . , vn−1 )], ρ0 (m)(v1 , . . . , vn−1 )i, if v1 , . . . , vn−1 are linearly independent and setting it equal to zero, if v1 , . . . , vn−1 are linearly dependent, where (x1 , . . . , xn ) is a coordinate system at m with ∂M described by xn = 0 and ρ(m)(v1 , . . . , vn−1 ) = hm, sign[µm ], ρ0 (m)(v1 , . . . , vn−1 )i with ρ0 (m) skew symmetric; moreover sign[µm ] = +1 (respectively, −1) if [µm ] and [σ(−∂/∂xn , v1 , . . . , vn−1 )] define the same (respectively, opposite) orientation of Tm M . If M = U , where U is open in Rn+ and ρ = α aξ0 ∈ Ωn−1 (U ), then τ i∗ ρ = (−1)n i∗ (aα)ξ0n−1 .

8.2 Stokes’ Theorem

469

In particular, if ζ=

n X

ζi dx1 ∧ · · · ∧ (dxi )ˆ ∧ · · · ∧ dxn ,

i=1

we have (i∗ ρ)(x1 , . . . , xn−1 ) = (−1)n a(x1 , . . . , xn−1 , 0)(−1)n−1 αn (x1 , . . . , xn−1 , 0) dx1 ∧ · · · ∧ (dxi )ˆ ∧ · · · ∧ dxn ξ0n−1 = −a(x1 , . . . , xn−1 , 0)αn (x1 , . . . , x−1 , 0) dx1 ∧ · · · ∧ (dxi )ˆ ∧ · · · ∧ dxn ξ0n−1 . With this observation, the proof of Stokes’ theorem 8.2.8 gives the following. 8.2.15 Theorem (Nonorientable Stokes’ Theorem). Let M be a paracompact nonorientable n-manifold with smooth boundary ∂M and ρ ∈ Ωn−1 (M ), a twisted (n − 1)-form with compact support. Then τ Z

Z

i∗ ρ.

dρ = M

∂M

The same statement holds for vector-valued twisted (n − 1)-forms and all corollaries go through replacing everywhere (n − 1)-forms with twisted (n − 1)-forms. For example, we have the following. 8.2.16 Theorem (Nonorientable Gauss Theorem). Let M be a nonorientable Riemannian n-manifold with associated density µM . Then for X ∈ X(M ) with compact support Z Z div(X)µM = (X · n)µ∂M M

∂M

where n is the outward unit normal of ∂M , µ∂M is the induced Riemannian density of ∂M and £X µM = (div X)µM . For a concrete situation in R3 involving these ideas, see Exercise 8.3-9.

Supplement 8.2B Stokes’ Theorem on Manifolds with Piecewise Smooth Boundary The statement of Stokes’ theorem we have given does not apply when M is, say a cube or a cone, since these sets do not have a smooth boundary. If the singular portion of the boundary (the four vertices and 12 edges in case of the cube, the vertex and the base circle in case of the cone), is of Lebesgue measure zero (within the boundary) it should not contribute to the boundary integral and we can hope that Stokes’ theorem still holds. This supplement discusses such a version of Stokes’ theorem inspired by Holmann and Rummler [1972]. (See Lang [1972] for an alternative approach.) First we shall give the definition of a manifold with piecewise smooth boundary. A glance at the definition of a manifold with boundary makes it clear that one could define a manifold with corners, by choosing charts that make regions near the boundary diffeomorphic to open subsets of a finite intersection of positive closed half-spaces. Unfortunately, singular points on the boundary—such as the vertex of a cone—need not be of this type. Thus, instead of trying to classify the singular points up to diffeomorphism and then make a formal intrinsic definition, it is simpler to consider manifolds already embedded in a bigger manifold. Then we can impose a condition on the boundary to insure the validity of Stokes’ theorem.

470

8. Integration on Manifolds

8.2.17 Definition. Definition Let U ⊂ Rn−1 be open and f : U → R be continuous. A point p on the graph of f , Γf = { (x, f (x)) | x ∈ U }, is called regular if there is an open neighborhood V of p such that V ∩ Γf is an (n − 1) dimensional smooth submanifold of V . Let ρf denote the set of regular points. Any point in σf = Γf \ρf is called singular . The mapping f is called piecewise smooth if ρf is Lebesgue measurable, π(σf ) has measure zero in U (where π : U × R → U is the projection) and f |π(σf ) is locally H¨ older; that is, for each compact set K ⊂ π(σf ) there are constants c(K) > 0, 0 < α(K) ≤ 1 such that |f (x) − f (y)| ≤ c(K)kx − ykα(K) for all x, y ∈ K. Note that ρf is open in Γf and that Int(Γ− f ) ∪ ρf , where Γ− f = { (x, y) ∈ U × R | y ≤ f (x) }, is a manifold with boundary ρf . Thus ρf has positive orientation induced from the standard orientation of Rn . This will be called the positive orientation of Γf . We are now ready to define manifolds with piecewise smooth boundary. 8.2.18 Definition. Let M be an n-manifold. A closed subset N of M is said to be a manifold with piecewise smooth boundary if for every p ∈ N there exists a chart (U, ϕ) of M at p, ϕ(U ) = U 0 × U 00 ⊂ Rn−1 × R, and a piecewise smooth mapping f : U 0 → R such that ϕ(bd(N ) ∩ U ) = Γf ∩ ϕ(U ) and ϕ(N ∩ U ) = Γ− f ∩ ϕ(U ). See Figure 8.2.7.

N

U M Figure 8.2.7. Singular boundary charts

It is readily verified that the condition on N is chart independent, using the fact that the composition of a piecewise smooth map with a diffeomorphism is still piecewise smooth. Thus, regular and singular points of bd(N ) make intrinsic sense and are defined in terms of an arbitrary chart satisfying the conditions of the preceding definition. Let ρN and σN denote the regular and singular part of the boundary bd(N ) of N in M. R To formulate Stokes’ theorem, we define N η, for η an n-form (respectively, density) on M with compact support. This is done as usual via a partition of unity; ρN and σN play no role since they have Lebesgue measure zero in every chart: ρN R because it is an (n − 1)-manifold and σN by definition. It is not so simple to define bd(N ) ζ for ζ ∈ Ωn−1 (M ) (respectively, a density). First a lemma is needed.

8.2 Stokes’ Theorem

471

8.2.19 Lemma. Let ζ ∈ Ωn−1 (U × R), where U is open in Rn−1 , supp(ζ) is compact and f : U → R is a piecewise smooth mapping. Then there is a smooth bounded function a : ρf → R, such that i∗ ζ = aλ where i : ρf → U × R is the inclusion and λ ∈ Ωn−1 (ρf ) is the boundary volume form induced by the canonical volume form of U × R ⊂ Rn on Int(Γ− f ) ∪ ρf . Proof. The existence of the function a on ρf follows since Ωn−1 (ρf ) is one-dimensional with a basis element λ. We prove that a is bounded. Let p ∈ ρf and v1 , . . . , vn−1 ∈ Tp (ρf ) be an orthonormal basis with respect to the Riemannian metric on ρf induced from the standard metric of Rn , and denote by n the outward unit normal. Then a(p) = a(p)(dx1 ∧ · · · ∧ dxn )(p)(n, v1 , . . . , vn−1 ) = a(p)λ(p)(v1 , . . . , vn−1 ) = ζ(p)(v1 , . . . , vn−1 ). Let vi = Since

vij

∂ ∂ ,..., ∂x1 p ∂xn p

∂ . ∂xj p and n, v1 , . . . , vn−1

are orthonormal bases of Tp (U × R), we must have |vij | ≤ 1 for all i, j. Hence if ζ=

n X

ζi dx1 ∧ · · · ∧ (dxi )ˆ ∧ · · · ∧ dxn ,

i=1

then |a(p)| = |ζ(p)(v1 , . . . , vn−1 )| X X n σ(1) σ(n−1) i = ζi (p) (−1) (sign σ)v1 · · · vn−1 i=1 σ∈Sn−1 =

n X

|ζi (p)|(n − 1)!

i=1

which is bounded, since ζ has compact support.

H

In view of this lemma and the fact that σf has measure zero, we can define Z Z Z ∗ ζ= i ζ= aλ. Γf

ρf

ρf

Now we can define, via a partition of unity, the integral of η ∈ Ωn−1 (M ) (or a twisted (n − 1)-form) by Z Z η= η. bd(N )

ρN

8.2.20 Theorem (Piecewise Smooth Stokes Theorem). Let M be a paracompact n-manifold and N a closed submanifold of M with piecewise smooth boundary. If (i) M is orientable and ω ∈ Ωn−1 (M ) has compact support, or (ii) M is nonorientable and ω ∈ Ωn−1 (M ) is a twisted (n − 1)-form (see the preceding supplement) which τ has compact support, then

472

8. Integration on Manifolds

Z

Z dω =

N

ω. bd(N )

The proof of this theorem reduces via a partition of unity to the local case. Thus it suffices to prove that if U is open in Rn−1 , ω ∈ Ωn−1 (U × R) has compact support, and f : U → R is a piecewise smooth mapping, then Z Z dω = ω. (8.2.11) Γ− f

Γf

The left-hand side of equation (8.2.11) is to be understood as the integral over the compact measurable set Γ− f ∩ supp(ω). For the proof of (8.2.11) we use three lemmas. 8.2.21 Lemma.

Equation (8.2.11) holds if ω vanishes in a neighborhood of σf in U × R.

Proof. Let V be an open neighborhood of σf in U × R on which ω vanishes and let W be another open neighborhood of σf (which is closed in V ) such that cl(W )∩(U ×R) ⊂ V . The set O = (U ×R)\ cl(W ) is open − and since it is disjoint from σf , Γ− f ⊂ O is an n-dimensional submanifold of O with bd(Γf ∩ O) = Γf ∩ O. Since − supp(dω) ∩ Γ− supp(ω) ∩ Γf ⊂ Γf ∩ O, f ⊂ Γf ∩ O and by the usual Stokes theorem, we have Z

Z

Z

dω =

dω = Γ− f ∩O

Γf

Z ω=

Γf ∩O

ω.



Γf

The purpose of the next two lemmas is to construct approximations to dω and ω if ω does not vanish near σf . For this we need translates of bump functions with control on their derivatives. 8.2.22 Lemma. Let C be a box (rectangular parallelepiped ) in Rn of edge lengths 2li and let D be the box with the same center as C but of edge lengths 4li /3. There exists a C ∞ function ϕ : Rn → [0, 1] which is 1 on Rn \C, 0 on D and |∂ϕ/∂xi | ≤ A/li , for a constant A independent of li . Proof. Assume we have found such a function ϕ : R → [0, 1] for n = 1. Then ψ(x1 , . . . , xn ) = ϕ(x1 ) . . . ϕ(xn ) is the desired function. The function ϕ is found in the following way. Let a = 2l/3,  = l/3 and choose an integer N such that 2/N < . Let h : R → [0, 1] be a bump function that is equal to 1 for |t| < 1/2 and that vanishes for |t| > 1. Then f : R → [0, 1], defined by f (t) = 1 − h(t) is a C ∞ function vanishing for |t| < 1/2 and equal to 1 for |t| > 1. Let fn (t) = f (nt) for all positive integers n and note that |fn0 (t)| = n|f 0 (nt)| ≤ Cn. Define the C ∞ function

 z  , fN t − 2N where the product is taken over integers z such that |z| < 2N a + 1. Note that if |t| < a + 1/4N and z ∈ Z is chosen such that |t − z/2N | < 1/4N , then  z  1 fN t − = 0 and |z| ≤ 2N |t| + < 2N a + 1, 2N 2 ϕ(t) =

Y

so that ϕ(t) = 0 . Similarly if |t| > a + then

2 N

and |z| < 2N a + 1,

z |z| 1 > t − ≥ |t| − 2N 2N N

8.2 Stokes’ Theorem

473

so that ϕ(t) = 1. Finally, let |t0 − a| < 2/N and let z0 ∈ Z be such that |t0 − z0 /2N | < 1/N . All factors fN (t0 − z/2N ) are one in a neighborhood of t0 , unless |t0 − z/2N | ≤ 1/N . In that case we have the inequality |z − z0 | ≤ |z − 2N t0 | + |2N t0 − z0 | ≤ 3. Hence at most seven factors in the product are not identically 1 in a neighborhood of t0 . Hence |ϕ0 (t0 )| ≤ 7CN =

A . 



8.2.23 Lemma. Let K be a compact subset of σf , the singular set of f . For every  > 0 there is a neighborhood U of K in U × R and a C ∞ function ϕ : U × R → [0, 1], which vanishes on a neighborhood of K in U , is one on the complement of U , and is such that   ∂ϕ (x) ≤ , i = 1, . . . , n, and (i) vol(U ) sup ∂xi x∈Rn (ii) vol(U ) ≤  and q(U ∩ ρf ) ≤ , where q is the measure on ρf associated with the volume form λ ∈ Ωn−1 (ρf ), and vol(U ) is the Lebesgue measure of U in Rn . Proof. Partition Rn−1 by closed cubes D of edge length 4l/3, l ≤ 1. At most 2n such cubes can meet at a vertex. The set π(K), where π : U × R → U is the projection, can be covered by finitely many open cubes C of edge length 2l, each one of these cubes containing a cube D and having the same center as C. Since π(K) and K have measure zero, choose l so small that for given δ > 0, S (i) the (n − 1)-dimensional volume of i=1,...,L Ci is smaller than or equal to δ; and  S   (ii) q π −1 C ∩ ρ ≤ δ. i f i=1,...,L Since f is locally H¨ older and π(K) is compact, there exist constants 0 < α ≤ 1 and k > 0 such that |f (x) − f (y)| ≤ kkx − ykα for x, y ∈ π(K). We can assume k ≥ 1 without loss of generality. In each of the sets π −1 (Ci ) = Ci × R, choose a box Pi with base Ci and height (2kl)1/α such that π(K) is covered by parallelopipeds Pi0 with the same centerSas Pi and edge lengths equal S to two-thirds of the edge lengths of Pi . Let V = i=1,...,L Pi . Then π(V ) = i=1,...,L Ci and since at most 2n of the Pi intersect vol(V ) = 2kl2n vol(π(V )) ≤ 2n+1 klδ ≤ 2n+1 kδ and q(V ∩ ρf ) ≤ δ. By the previous lemma, for each Pi there is a C ∞ function ϕi : U × R → [0, 1] that vanishes on Pi0 , is equal to 1 on the complement of Pi , and

∂ϕi A

sup j

≤ l. x∈Rn ∂x Let ϕ = Πi=1,...,L ϕi . Clearly ϕ : U × R → [0, 1] is C ∞ , vanishes in a neighborhood of K and equals one in the complement of V . But at most 2n of the Pi can intersect, so that L Y ∂ϕ X ∂ϕi = ≤ 2n A , j = 1, . . . , n. ϕ k ∂xj j ∂x l i=1 k6=i

474

8. Integration on Manifolds

Hence  ∂ϕ A vol(V ) sup j ≤ 2n+1 klδ2n = 22n+1 kδA. l x∈Rn ∂x 

Now let δ = min{, /22n+1 kA}, ϕ = ϕ, and U = V. Proof of Equation (8.2.11). ω=

n X



Let

ω i dx1 ∧ · · · ∧ (dx1 )ˆ ∧ · · · ∧ dxn ,

dω = b dx1 ∧ · · · ∧ dxn ,

i=1

and i∗ ω = aλ. Then ω i , b, and a are continuous and bounded on U × R and ρf respectively; that is, |ω i (x)| ≤ M , |b(x)| ≤ N for x ∈ U × R and |a(y)| ≤ N for y ∈ ρf , where M, N > 0 are constants. Let U and ϕ be given by the previous lemma applied to supp(ω) ∩ σf . But ϕ ω vanishes in a neighborhood of σf and Lemma 8.2.21 is applicable; that is Z Z d(ϕ ω) = ϕ ω. (8.2.12) Γ− f

Γf

We have Z Z Z a(1 − ϕ )λ ≤ N q(U ∩ ρf ) ≤ N  ω− ϕ ω ≤ ρf Γf Γf and Z Z Z Z dω − d(ϕ ω) ≤ (dω − ϕ dω) + dϕ ∧ ω − − − Γ− Γf Γf Γf f Z n Z X ∂ϕ |ω i | i dx1 ∧ · · · ∧ dxn b(1 − ϕ )dx1 ∧ · · · ∧ dxn + ≤ − Γ− ∂x f i=1 Γf " n # X ∂ϕ (x) vol(U ) ≤ N  + M n. sup ≤ N vol(U ) + M i ∂x x∈R i=1 From equations (8.2.12) and (8.2.13) we get Z Z dω − Γ− Γf f

for all  > 0, which proves the equality.

(8.2.13)

ω ≤ (2N + nM ) 

In analysis it can be useful to have hypotheses on the smoothness of ω as well as on the boundary that are as weak as possible. Our proofs show that ω need only be C 1 . An effective strategy for sharper results is to approximate ω by smooth forms ωk so that both sides of Stokes’ theorem converge as k → ∞. A useful class of forms for which this works are those in Sobolev spaces, function spaces encountered in the study of partial differential equations. The H¨ older nature of the boundary of N in Stokes’ theorem is exactly what is needed to make this approximation process work. The key ingredients are approximation properties in M (which are obtained from those in Rn ) and the Calder´ on extension theorem to reduce approximations in N to those in Rn . (Proofs of these facts may be found in Stein [1970], Marsden [1973], and Adams [1975].)

8.2 Stokes’ Theorem

475

Supplement 8.2C Stokes’ Theorem on Chains In algebraic topology it is of interest to integrate forms over images of simplexes. This box adapts Stokes’ theorem to this case. The result could be obtained as a corollary of the piecewise smooth Stokes Theorem, but we shall give a self-contained and independent proof. 8.2.24 Definition.

The standard p-simplex is the closed set ) ( p X i i p x ≤1 . ∆p = x ∈ R 0 ≤ x ≤ 1, i=1

The vertices of ∆p are the p + 1 points v0 = (0, . . . , 0), v1 = (1, 0, . . . , 0), . . . , vp = (0, . . . , 0, 1). Opposite to each vi there is the ith face Φp−1,i : ∆p−1 → ∆p given by (see Figure 8.2.8): ! p−1 X 1 p−1 i 1 p−1 Φp−1,0 (y , . . . , y )= 1− y ,y ,...,y , if i = 0 i=1

and Φp−1,i (y 1 , . . . , y p−1 ) = (y 1 , . . . , y i−1 , 0, y i , . . . , y p−1 ), k

r

if i 6= 0.

k

A C -singular p-simplex on a C -manifold M , 1 ≤ k ≤ r, is a C -map s : U → M , where U is an open neighborhood of ∆p in Rp . The points s(v0 ), . . . , s(vp ) are the vertices of the singular p-simplex s and the map s ◦ Φp−1,i : V → M , for V an open neighborhood of ∆p−1 in Rp−1 and Φp−1,i extended by the same formula from ∆p−1 to V , is called the ith face of the singular p-simplex s. A C k -singular p-chain on M is a finite formal linear combination with real coefficients of C k -singular p-simplexes. The boundary of a singular p-simplex s is the singular (p − 1)-chain ∂s defined by ∂s =

p X

(−1)i s ◦ Φp−1,i

i=0

and that of a singular p-chain is obtained by extending ∂ from the simplexes by linearity to chains. It is straightforward to verify that ∂ ◦ ∂ = 0 using the relation Φp−1,j ◦ Φp−2,j = Φp−2,i−1 ◦ Φp−2,i−1 for j < i. If s : U → M, ∆p ⊂ U , is a singular p-simplex, ω ∈ Ωp (M ), and s∗ ω = adx1 ∧ · · · ∧ dxp ∈ Ωp (U ), the integral of ω over s is defined by Z

Z ω=

s

a dx1 · · · dxp ,

∆p

where the integral on the right is the usual integral in Rp . The integral of ω over a p-chain is obtained by linear extension.

476

8. Integration on Manifolds

Figure 8.2.8. Integrating over chains

If c is any singular p-chain and ω ∈ Ωp−1 (M ), then

8.2.25 Theorem (Stokes’ Theorem on Chains). Z

Z dω =

ω.

c

Proof.

∂c

By linearity it suffices to prove the formula if c = s, a singular p-simplex. If s∗ ω =

p X

(−1)j−1 ω j dx1 ∧ · · · ∧ (dxj )ˆ ∧ · · · ∧ dxp ,

j=1

then d(s∗ ω) =

p−1 X ∂ω j j=1

∂xj

dx1 ∧ · · · ∧ dxp

and denoting the coordinates in a neighborhood V of ∆p−1 by (y 1 , . . . , y p−1 ) = y, we get Φ∗p−1,0 s∗ ω(y)

=

p X

ω

j

1−

j=1

p−1 X

! i

1

y ,y ,...,y

p−1

dy 1 ∧ · · · ∧ dy p−1 ,

i=1

if i = 0 and Φ∗p−1,i s∗ ω(y) = (−1)i−1 ω i (y 1 , . . . , y i−1 , 0, y i , . . . , y p−1 ) dy 1 ∧ · · · ∧ dy p−1 , if i 6= 0. Thus, the formula in the statement becomes p Z X j=1

∂ω j (x) 1 dx · · · dxp ∂xj ∆p " ! p Z p−1 X X j i 1 p−1 = ω 1− y ,y ,...y j=1

∆p−1

(8.2.14)

i=1

# j

1

− ω (y , . . . , y

j−1

j

, 0, y , . . . , y

p−1

) dy 1 · · · dy p−1 .

8.2 Stokes’ Theorem

477

By Fubini’s theorem, each summand on the left hand side of equation (8.2.14) equals Z ∆p

∂ω j (x) 1 dx · · · dxp ∂xj ! Z Z 1−Pk6=j xk ∂ω j j = dx dx1 · · · (dxj )∧ · · · dxp ∂xj ∆p−1 0    Z X ω j x1 , . . . , xj−1 , 1 − = xk , xj+1 , . . . , xp  ∆p−1

k6=j

# j

1

j−1

− ω (x , . . . , x

j+1

, 0, x

, . . . , x ) dx1 · · · (dxj )∧ · · · dxp . p

Break up this integral as a difference of two terms. In the first term perform the change of variables   X (y 1 , . . . , y p−1 ) 7→ x2 , . . . , xj−1 , 1 − xk , xj+1 , . . . , xp  k6=j

which has Jacobian equal to (−1)j , use the change of variables formula from calculus in the multiple integral involving the absolute value of the Jacobian, and note that x1 = 1 − Σi=1,...,p−1 y i . In the second term perform the change of variables (y 1 , . . . , y p−1 ) 7→ (x1 , . . . , xj−1 , xj+1 , . . . , xp ) which has Jacobian equal to one. Then we get Z ∂ω j (x) 1 dx . . . dxp j ∂x ∆p " ! Z p−1 X = ωj 1 − y i , y 1 , . . . , y p−1 ∆p−1

i=1

# j

1

− ω (y , . . . , y

j−1

j

, 0, y , . . . , y

p−1

and formula (8.2.14) is thus proved for each corresponding summand.

) dy 1 . . . dy p−1 

Instead of singular p-chains one can consider infinite singular p-chains defined as infinite formal sums P with real coefficients i∈I ai Si such that for each i ∈ I the family of sets { si (∆p ) | ai 6= 0 } is locally finite, that is, each m ∈ M has a neighborhood intersecting only finitely many (or no) sets of this family. On compact manifolds only finitely many coefficients in an infinite singular p-chain are non-zero and thus infinite singular p-chains are singular p-chains. The statement and the proof of Stokes’ theorem on chains remain unchanged if c is an infinite p-chain and ω ∈ Ωπ−1 (M ) has compact support.

Exercises  8.2-1. Let M and N be oriented n-manifolds with boundary and f : M → N an orientation-preserving diffeomorphism. Show that the change of variables formula and Stokes’ theorem imply that f ∗ ◦ d = d ◦ f ∗ .

478

8. Integration on Manifolds

 8.2-2. Let M be a compact orientable boundaryless n-manifold and α ∈ Ωn−1 (M ). Show that dα vanishes at some point.  8.2-3. Let M be a compact oriented (n + 1)-dimensional manifold with boundary, R f : ∂M → N a smooth map and ω ∈ Ωn (N ) where dω = 0. Show that if f extends smoothly to M , then ∂M f ∗ ω = 0.  8.2-4.

Let (M, µ) be a volume manifold with ∂M = ∅.

(i) Show that the divergence of a vector field X is uniquely determined by the condition Z Z f (div X)µ = − (£X f )µ M

M

for any f with compact support. (ii) What does the equation in (i) become if M is compact with boundary? (iii) X(x, y, z) = (y, −x, 0) defines a vector field on S 2 . Calculate div X.  8.2-5. Let M be a paracompact manifold with boundary. Show that there is a positive smooth function f : M → [0, ∞[ with 0 a regular value, such that ∂M = f −1 (0). Hint: First do it locally and then patch the local functions together with a partition of unity.  8.2-6. Let M be a boundaryless manifold and f : M → R a C ∞ mapping having a regular value a. Show that f −1 ([a, ∞[) is a manifold with boundary f −1 (a).  8.2-7. Let f : M → N be a C ∞ mapping, ∂M 6= ∅, ∂N 6= ∅, and let P ⊂ N be a submanifold of N . Assume that f t P , (f |∂M ) t P and that in addition one of the following conditions hold. (i) P is boundaryless and P ⊂ Int N ; or (ii) ∂P 6= ∅ and ∂P ⊂ ∂N ; or (iii) ∂P 6= ∅, f t ∂P , and (f |∂M ) t ∂P . Show that f −1 (P ) is a submanifold of M whose boundary equals ∂f −1 (P ) = f −1 (P ) ∩ ∂M, in case (i), and ∂f −1 (P ) = f −1 (∂P ) in cases (ii) and (iii). If all manifolds are finite dimensional, show that dim M − dim f −1 (P ) = dim N − dim P. Formulate and prove the statement replacing this equality between dimensions for infinite dimensional manifolds. Hint: At the boundary, work with a boundary chart using the technique in the proof of Theorem 3.5.12.  8.2-8. Without some kind of transversality conditions on f |∂M for f : M → N a smooth map, even if f t P , where P is a submanifold of N (like the ones in the previous exercise), f −1 (P ) is in general not a submanifold. For example, let M = R2+ , N = R, P = {0} and f (x, y) = y + χ(x) for a smooth function χ : R → R . Show that f is a smooth surjective submersion. Find the conditions under which f |∂M has 0 as a regular value. Construct a smooth function χ for which these conditions are violated and f −1 (0) is not a manifold. Hint: Take for χ a smooth function which has infinitely many zeros converging to zero.  8.2-9.

(i) Show that if M is a boundaryless manifold, there is a connected manifold N with M = ∂N .

Hint: Think of semi-infinite cylinders. (ii) Construct an example for (i) in which M is compact but N cannot be chosen to be compact. Hint: Assume dim M = 0.  8.2-10. Let M be a manifold, X a smooth vector field on M with flow Ft and α ∈ Ωk (M ). We call α an invariant k-form of X when £X α = 0. Prove the following.

8.2 Stokes’ Theorem

479

Poincar´ e–Cartan Theorem. α is an invariant k-form of X iff for all oriented compact k-manifolds with boundary (V, ∂V ) and C ∞ mappings ϕ : V → M , such that the domain of Ft contains ϕ(V ), 0 ≤ t ≤ T , we have Z Z (Ft ◦ ϕ)∗ α =

V

ϕ∗ α.

V

Hint: For the converse show that the equality between integrals implies (Ft ◦ ϕ)∗ α = ϕ∗ α; then differentiate relative to t.  8.2-11. Let X be a vector field on a manifold M and α, β invariant forms of X. (See Exercise 8.2-10.) Prove the following. (i) iX α is an invariant form of X. (ii) dα is an invariant form of X. (iii) £X γ is closed iff dγ is an invariant form, for any γ ∈ Ωk (M ). (iv) α ∧ β is an invariant form of X. (v) Let AX denote the invariant forms of X. Then AX is a ∧ subalgebra of Ω(M ), which is closed under d and iX .  8.2-12. Let X be a vector field on a manifold M with flow Ft and α ∈ Ωk (M ). Then α is called a relatively invariant k-form of X if £X α is closed. Prove the following Poincar´ e–Cartan Theorem. α is a relatively invariant (k − 1)-form of X iff for all oriented compact k-manifolds with boundary (V, ∂V ) and C ∞ maps ϕ : V → M such that the domain of Ft contains ϕ(V ) for 0 ≤ t ≤ T , we have Z Z (Ft ◦ ϕ ◦ i)∗ α =

∂V

(ϕ ◦ i)∗ α.

∂V

where i : ∂V → V is the inclusion map.  8.2-13. If X ∈ X(M ), let AX be the set of all invariant forms of X, RX the set of all relatively invariant forms of X, C the set of all closed forms in Ω(M ), and E the set of all exact forms in Ω(M ). Show that (i) AX ⊂ RX , E ⊂ C ⊂ RX , AX is a differential subalgebra of Ω(M ), but RX is only a real vector subspace. £

i

π

X (ii) 0 → AX → Ω(M ) → Ω(M ) → Ω(M )/ Im(£X ) → 0 is exact.

i

d

π

(iii) 0 → C → RX → AX → AX /E ∩ AX → 0 is exact. (iv) d(AX ) ⊂ AX and iX (AX ) ⊂ AX .  8.2-14 (Smale–Sard Theorem for manifolds with boundary). Let M and N be C k manifolds, where M is Lindel¨of, having a boundary ∂M , and N is boundaryless. Let f : M → N be a C k Fredholm map and let ∂f = f |∂M . If k > max(0, index(Tx f )) for every x ∈ M , show that Rf ∩ R∂f is residual in N .  8.2-15 (The Boundaryless Double). Let M be a manifold with boundary. Show that the topological space obtained by identifying the points of ∂M in the disjoint union of M with itself is a boundaryless manifold in which M embeds, called the boundaryless double of M . Hint: Glue together the two boundary charts.

480

8. Integration on Manifolds

 8.2-16. Let M be a manifold with ∂M 6= ∅. Assume M admits partitions of unity. Show that ∂M is orientable and hence by Exercise 7.5-17, the algebraic normal bundle ν(∂M ) = (T M |∂M )/T (∂M ) is trivial. Hint: Use proposition 7.5.8. Locally n(m) = ∂/∂xn for m ∈ ∂M ; glue these together.  8.2-17 (Collars). Let M be a manifold with boundary. A collar for M is a diffeomorphism of ∂M × [0, 1[ onto an open neighborhood of ∂M in M that is the identity on ∂M. (i) Show that a manifold with boundary and admitting partitions of unity has a collar. Hint: Via a partition of unity, construct a vector field on M that points inward when restricted to ∂M. Then look at the integral curves starting on ∂M to define the collar. (ii) Let ϕ1 : ∂M × [0, 1[ → M , i = 0, 1 be two collars. Show that ϕ0 and ϕ1 are isotopic, that is, there is a smooth map H : ]−, 1 + [ × ∂M × [0, 1[ → M such that H(0, m, t) = ϕ0 (m, t),

H(1, m, t) = ϕ1 (m, t)

for all (m, t) ∈ ∂M × [0, 1[ and that H(s, ·, ·) is an embedding for all s ∈ ]−, 1 + [.2 Hint: Let Ui be the image of ϕi , an open set in M containing ∂M . Let Xi = ϕ∗i (0, ∂/∂t) and look at the flow of (1 − s)X0 + sX1 on U0 ∩ U1 . (iii) Let N be a submanifold of M such that ∂N = N ∩ ∂M and Tn N is not a subset of Tn (∂M ) for all n ∈ ∂N . Show that ∂M has a collar which restricts to a collar of ∂N in N .  8.2-18. Let M and N be manifolds with boundary and let ϕ : ∂M → ∂N be a diffeomorphism. Form the topological space M ∪ϕ N which is the quotient of the disjoint union of M with N by the equivalence relation which identifies m with ϕ(m). Let V be the image of ∂M and ∂N in M ∪ϕ N . (i) Use collars to construct a homeomorphism of a neighborhood U of V with the space ]−1, 1[ × V which maps V pointwise to V × {0} and which maps V ∩ M and V ∩ N diffeomorphically onto V × ]0, 1[ and V × ]−1, 0[, respectively. Construct a differentiable structure out of those on M , N, and U . The “uniqueness theorem of glueing” states that the differentiable structures on the space M ∪ϕ N obtained in (i) by making various choices are all diffeomorphic. The rest of this exercise uses this fact. (ii) Two compact boundaryless manifolds M1 , M2 are called cobordant if there is a compact manifold with boundary N , called the cobordism, such that ∂N equals the disjoint union of M1 with M2 . Show that “cobordism” is an equivalence relation. Hint: For transitivity, glue the manifolds along one common component of their boundaries. (iii) Show that the operation of disjoint union induces the structure of an abelian group on the set of cobordism classes in which each element has order two. Hint: The zero element is the class of any compact manifold which is the boundary of another compact manifold. (iv) Show that the operation of taking products of manifolds induces a multiplicative law on the set of cobordism classes, thus making this set N a ring. 2 It

can be shown that ϕ0 and ϕ1 are diffeotopic using Thom’s theorem of embedding of isotopies into diffeotopies. This then provides the basis of glueing manifolds together along their boundaries; see Hirsch [1976, Chapter 8], for proofs and the preamble to the next exercise for a discussion.

8.2 Stokes’ Theorem

481

(v) Repeat parts (ii) and (iii) for oriented manifolds obtaining a graded ring, that is, [M, µ] · [N, ν] = (−1)dim M +dim N [N, ν] · [M, µ], the graded ring of oriented cobordism classes O. Are the elements of O still of order two relative to addition? (vi) Denote by Nn , On , the cobordism classes of a given dimension. Show that N0 = Z/2Z, O0 = Z, N1 = O1 = 0. (vii) Assume M and N are boundaryless manifolds, M compact, and P a boundaryless submanifold of N . Assume f, g : M → N are smoothly homotopic maps such that f t P and g t P . Show that f −1 (P ) and g −1 (P ) are cobordant. Hint: Choose a smooth homotopy H transverse to P . What is ∂H −1 (P )?  8.2-19.

This exercise concerns vector and tensor field densities.

(i) Let χ be a vector field density on a finite dimensional manifold M , that is, χ = X ⊗ρ for X ∈ X(M ) and ρ ∈ |Ω(M )|. Recall from Exercise 7.5-16 that the density div χ, defined to be (divρ X)ρ, is independent of the representation of χ as X ⊗ ρ. Show that for any f ∈ F(M ), Y ∈ X(M ), and any diffeomorphism ϕ : M → M we have ϕ∗ (div χ) = div(∗ χ), £Y (div χ) = div(£Y χ), and div(f χ) = df · χ + f div χ. (ii) If M is paracompact and Riemannian, phrase Gauss’ theorem for vector field densities. (iii) Let α ∈ Ωk−1 (M ) and τ be a tensor density of type (k, 0) which is completely antisymmetric. Let α · τ denote the contraction of α with the first k − 1 indices of the tensor part of τ producing a vector field density. Define the contravariant exterior derivative ∂τ by requiring the following relation for all α ∈ Ωk−1 (M ) : div(α · τ ) = dα · τ + α · ∂τ, where dα · τ and α · ∂τ means contraction on all indices. Show that if τ = t ⊗ ρ, where locally t = ti1 ...ik

∂ ∂ ∧ ··· ∧ i , ∂xi1 ∂x k

and ρ = |dx1 ∧ · · · ∧ dxn |,

then the local expression of ∂t is ∂τ =

∂ i1 ...ik−1 j ∂ ∂ (t ) i1 ∧ · · · ∧ ik−1 ⊗ |dx1 ∧ · · · ∧ dxn |. ∂xj ∂x ∂x

Show that ∂ 2 = 0. (iv) Prove the following properties of ∂: ∂(τ ∧ σ) = ∂τ ∧ σ + (−1)k τ ∧ ∂σ, and £X ∂τ = ∂£X τ,

ϕ∗ ∂τ = ∂ϕ∗ τ

where τ is a (k, 0)-tensor density, σ is a (l, 0)-tensor density, X ∈ X(M ), and ϕ : M → M is a diffeomorphism. Show that if χ is a vector field density, ∂χ = div χ.

482

8. Integration on Manifolds

(v) Let jX τ = X ∧ τ for X ∈ X(M ). Show that iX α · τ = α · jX τ for any α ∈ Ωk+1 (M ), X ∈ X(M ) and τ is a completely antisymmetric (k, 0)-tensor density. Prove the analog of Cartan’s formula: £X = jX ◦ ∂ + ∂ ◦ jX . Hint: Integrate the defining relation in (iii) for a any form with support in Int M and extend the formula by continuity to ∂M . (vi) Formulate and prove a global formula for ∂τ (α1 , . . . , αk−1 ), τ a (k, 0)-tensor density, analogous to Palais’ formula (Proposition 7.4.11(ii)).  8.2-20 (Pr¨ ufer Manifold). Let Rd denote the set R with the discrete topology; it is thus a zero dimensional manifold. Let P = (R2+ × Rd )/R, where R2+ = { (x, y) | y ≥ 0 } and R is the equivalence relation: (x, y, a) R(x0 , y 0 , a0 ) iff [(y = y 0 > 0 and a + xy = a0 + x0 y 0 ) or (y = y 0 = 0 and a = a0 , x = x0 )]. Show that P is a Hausdorff two-dimensional manifold, ∂P 6= ∅, and ∂P is a disjoint union of uncountably many copies of R. Show that P is not paracompact.  8.2-21.

8.3

In the notation of Supplement 8.2C, verify that ∂ ◦ ∂ = 0.

The Classical Theorems of Green, Gauss, and Stokes

This section obtains these three classical theorems as a consequence of Stokes’ theorem for differential forms. We begin with Green’s theorem, which relates a line integral along a closed piecewise smooth curve C in the plane R2 to a double integral over the region D enclosed by C. (Piecewise smooth means that the curve C has only finitely many corners.) Recall from advanced calculus that the line integral of a one-form ω = P dx + Q dy along a curve C parameterized by γ : [a, b] → R2 is defined by Z

Z ω=

C

a

{P (γ1 (t), γ2 (t))γ10 (t) + Q(γ1 (t), γ2 (t))γ20 (t)}dt;

b

that is, Z

Z

b

ω= C

γ ∗ ω.

a

8.3.1 Theorem (Green’s Theorem). Let D be a closed bounded region in R2 bounded by a closed positively oriented piecewise smooth curve C. (Positively oriented means the region D is on your left as you traverse the curve in the positive direction.) Suppose P : D → R and Q : D → R are C 1 . Then ZZ 

Z P dx + Q dy = C

D

∂Q ∂P − ∂x ∂y

 dx dy.

Proof. We assume the boundary C = ∂D is smooth. (The piecewise smooth case follows from the generalization of Stokes’ theorem outlined in Supplement 8.2B). Let ω = P (x, y)dx + Q(x, y)dy ∈ Ω1 (D). Since  dω =

∂Q ∂P − ∂x ∂y

 dx ∧ dy

and the measure associated with the volume dx ∧ dy on R2 is the usual Lebesgue measure dx dy, the formula of the theorem is a restatement of Stokes’ theorem for this case. 

8.3 The Classical Theorems of Green, Gauss, and Stokes

483

This theorem may be phrased in terms of the divergence and the outward unit normal. If C is given parametrically by t 7→ (x(t), y(t)), then the outward unit normal is (y 0 (t), −x0 (t)) n= p x0 (t)2 + y 0 (t)2

(8.3.1)

and the infinitesimal arc-length (the volume element of C) is p ds = x0 (t)2 + y 0 (t)2 d. (See Figure 8.3.1.) If X=P

∂ ∂ +Q ∈ X(D), ∂x ∂y

recall that div X = ∗d ∗ X[ = 8.3.2 Corollary. then

∂Q ∂P + . ∂x ∂y

Let D be a region in R2 bounded by a closed piecewise smooth curve C. If X ∈ X(D), Z ZZ (X · n)ds = (div X)dx dy. C

D

R

where C f ds denotes the line integral of the function f over the positively oriented curve C and X · n is the dot product.

D

C Figure 8.3.1. Green’s Theorem

Proof.

Using formula (8.3.1) for n, we have Z

Z

b

(X · n)ds = C

[P (x(t)y(t))y 0 (t) − Q(x(t), y(t))x0 (t)]dt

a

Z P dy − Q dx

=

(8.3.2)

C

by the definition of the line integral. By Theorem 8.3.1, equation (8.3.2) equals  ZZ  ZZ ∂P ∂Q + dx dy = (div X)dx dy. ∂x ∂y D D Taking P (x, y) = −y and Q(x, y) = x in Green’s theorem, we get the following.



484

8. Integration on Manifolds

8.3.3 Corollary. is given by

Let D be a region in R2 bounded by a closed piecewise smooth curve C. The area of D

1 area(D) = 2

Z x dy − y dx. C

The classical Stokes theorem for surfaces relates the line integral of a vector field around a simple closed curve C in R3 to an integral over a surface S for which C = ∂S. Recall from advanced calculus that the line integral of a vector field X in R3 over the curve σ : [a, b] → R3 is defined by Z Z b X · ds = X(σ(t)) · σ 0 (t)dt. (8.3.3) σ

a

The surface integral of a compactly supported two-form ω in R3 is defined to be the integral of the pullback of ω to the oriented surface. If S is an oriented surface, n is called the outward unit normal at x ∈ S if n is perpendicular to Tx S and {n, e1 , e2 } is a positively oriented basis of R3 whenever {e1 , e2 } is a positively oriented basis of Tx S. Thus S is orientable iff the normal bundle to S, which has one-dimensional fiber, is trivial. Also, the area element ν of S is given by Proposition 7.5.8. That is, ν(x)(v1 , v2 ) = µ(x)(n, v1 , v2 ), (8.3.4) R where v1 , v2 ∈ Tx S, and µ = dx ∧ dy ∧ dz. We want to express S ω in a form familiar from vector calculus. Let ω = P dy ∧ dz + Q dz ∧ dx + R dx ∧ dy so that ω = ∗X[ , where X=P

∂ ∂ ∂ +Q +R . ∂x ∂y ∂z

Recall that α ∧ ∗β = hα, βiµ so that letting α = n[ , and β = X[ , we get n[ ∧ ∗X[ = (X · n)µ. Applying both sides to (n, v1 , v2 ) and using equation (8.3.2) gives (n[ ∧ ∗X[ )(n, v1 , v2 ) = (X · n)ν(v1 , v2 )

(8.3.5)

(the base point x is suppressed). The left side of (8.3.5) is ∗X[ (v1 , v2 ) since n[ is one on n and zero on v1 and v2 . Thus (8.3.5) becomes ∗ X[ = (X · n)ν. (8.3.6) Therefore, Z

Z (X · n)dS =

ω= S

Z

S

X · dS, S

where dS, the measure on S defined by ν, is identified with a surface integral familiar from vector calculus. R A physical interpretation of S (X · n)dS may be useful. Think of X as the velocity field of a fluid, so X is pointing in the direction in which the fluid is moving across the surface S and X · n measures Rthe volume of fluid passing through a unit square of the tangent plane to S in unit time. Hence the integral S (X · n)dS is the net quantity of fluid flowing across the surface per unit time, that is, the rate of fluid flow. Accordingly, this integral is also called the flux of X across the surface. 8.3.4 Theorem (Classical Stokes Theorem). Let S be an oriented compact surface in R3 and X a C 1 vector field defined on S and its boundary. Then Z Z (curl X) · n dS = X · ds. S

where n is the outward unit normal to S (Figure 8.3.2).

∂S

8.3 The Classical Theorems of Green, Gauss, and Stokes

485

Proof. FirstRextend X viaR a bump function to all of R3 so that the extended X still has compact support. By definition, ∂S X · ds = ∂S X[ where [ denotes the index lowering action defined by the standard metric in R3 . But dX[ = ∗(curl X)[ (see Example 7.4.3C) so that by equations (8.3.3), (8.3.6), and Stokes’ theorem, Z Z Z Z [ [ X · ds = dX = ∗(curl X) = (curl X · n)dS.  ∂S

S

S

S

X

S

Figure 8.3.2. Stokes’ Theorem

8.3.5 Examples. A. The historical origins of Stokes’ formula, are connected with Faraday’s law, which is discussed in Chapter 9 and example B below. In fluid dynamics, Stokes’ formula is useful in the development of Kelvin’s circulation theorem, to be discussed in §9.2. Here we concentrate on a physical interpretation of the curl operator. Suppose X represents the velocity vector field of a fluid. Let us apply Stokes’ theorem to a disk Dr of radius r at a point P ∈ R3 (Figure 8.3.3). We get Z Z X · ds = (curl X) · nds = (curl X · n)(Q)πr2 , ∂Dr

Dr

the last equality coming from the mean value theorem for integrals; here Q ∈ Dr is some point given by the mean value theorem and πr2 is the area of Dr . Thus Z 1 ((curl X) · n)(P ) = lim 2 X · ds. (8.3.7) r→0 πr ∂Dr R The number C X · ds is called the circulation of X around the closed curve C. It represents the net amount of turning of the fluid in a counterclockwise direction around C.

P r Figure 8.3.3. Curl is the circulation per unit area

Formula (8.3.7) gives the following physical interpretation for curl X, namely: (curl X)·n is the circulation of X per unit area on a surface perpendicular to n. The magnitude of (curl X) · n is clearly maximized when n = (curl X)/k curl Xk. The vector curl X is called the vorticity vector.

486

8. Integration on Manifolds

B. One of Maxwell’s equations of electromagnetic theory states that if E(x, y, z, t) and B(x, y, z) represent the electric and magnetic fields at time t, then ∇×E=−

∂B , ∂t

where ∇ × E is computed by holding t fixed, and ∂B/∂t is computed by holding x, y, and z constant. Let us use Stokes’ theorem to determine what this means physically. Assume S is a surface to which Stokes’ theorem applies. Then Z Z Z Z ∂ dB · dS = − B · dS. E · ds = (∇ × E) · dS = − ∂t S ∂S S S ∂t (The last equality may be justified if B is C 1 .) Thus we obtain Z Z ∂ E · ds = − B · dS. (8.3.8) ∂t S ∂S R Equality (8.3.8) is known as Faraday’s law. The quantity ∂S E · ds representsR the “voltage” around ∂S, and if ∂S were a wire, a current would flow in proportion to this voltage. Also S B · dS is called the flux of B, or the magnetic flux. Thus, Faraday’s law says that the voltage around a loop equals the negative of the rate of change of magnetic flux through the loop. C. Let X ∈ X(R3 ). Since R3 is contractible, the Poincar´e lemma shows that curl X = 0 iff X = grad f for some function f ∈ F(R3 ). This in turn R is equivalent (by Stokes’ theorem) to either of the following: (i) for any oriented simple closed curve C, C X · ds = 0, or (ii) for any oriented simple curves C1 , C2 with the same end points, Z Z X · ds.

X · ds = C1

C2

The function f can be found in the following way: Z x Z f (x, y, z) = X 1 (t, 0, 0)dt + 0

y

X 2 (x, t, 0)dt +

0

Z

z

X 3 (x, y, t)dt.

(8.3.9)

0

Thus, for example, if X=y

∂ ∂ ∂ + (z cos(yz) + x) + y cos(yz) , ∂x ∂y ∂z

then curl X = 0 and so X = grad f , for some f . Using the formula (8.3.9), one finds f (x, y, z) = xy + sin yz.

D.

(8.3.10)

The same arguments apply in R2 using Green’s theorem in place of Stokes’ theorem. Namely, if X = X1

∂ ∂ + X2 ∈ X(R2 ) ∂x ∂y

and

∂X 2 ∂X 1 = , ∂x ∂y

then X = grad f , for some f ∈ F(R2 ) and conversely. E. The following statement is again a reformulation of the Poincar´e lemma: let X ∈ X(R3 ), then div X = 0 iff X = curl Y for some Y ∈ X(R3 ). 

8.3 The Classical Theorems of Green, Gauss, and Stokes

487

Figure 8.3.4. Gauss’ Theorem

8.3.6 Theorem (Classical Gauss Theorem). Let Ω be a compact set with nonempty interior in R3 bounded by a surface S that is piecewise smooth. If X is a C 1 vector field on Ω ∪ S, then Z Z (div X)dV = (X · n)dS, (8.3.11) Ω

S

where dV denotes the standard volume element (Lebesgue measure) in R3 (Figure 8.3.4). Proof.

Either use Corollary 8.2.10 or argue as in Theorem 8.3.4. By equation (8.3.6), Z Z (X · n)dS = ∗X[ . S

S

By Stokes’ theorem, this equals Z

[

Z

d∗X = Ω

(div X)dV. Ω

since d ∗ X[ = (div X)dx ∧ dy ∧ dz. 8.3.7 Example.



We shall use the preceding theorem to prove Gauss’ law ( Z 4π, if 0 ∈ Ω; r·n ds = 3 0, if 0 ∈ / Ω, ∂Ω r

(8.3.12)

where Ω is a compact set in R3 with nonempty interior, ∂Ω is the surface bounding Ω, which is assumed to be piecewise smooth, n is the outward unit normal, 0 ∈ / ∂Ω, and where r = (x2 + y 2 + z 2 )1/2 ,

r = (x, y, z).

If 0 ∈ / Ω, apply Theorem 8.3.6 and the fact that div(r/r3 ) = 0 to get the result. If 0 ∈ Ω, surround 0 inside Ω by a ball D of radius  (Figure 8.3.5). Since the orientation of ∂D induced from Ω\D is the opposite of that induced from D (namely it is given by the inward unit normal), Gauss’ theorem gives Z Z Z r·n r·n r·n dS + dS = dS = 0 (8.3.13) 3 3 3 ∂Ω r ∂D r ∂(Ω\D ) r since 0 ∈ / ∂Ω\D and thus on Ω\D , we have div(r/r3 ) = 0. But on ∂D , r =  and n = −r/, so that Z Z r·n 2 1 dS = − dS = − 2 4π2 = −4π, 3 4  ∂D r ∂D 

488

8. Integration on Manifolds

since

Z

dS = 4π2 ,

∂D

the area of the sphere of radius .



Figure 8.3.5. The regions used to prove Gauss’ Law.

In electrostatics Gauss’ law (8.3.12) is used in the following way. The potential due to a point charge q at 0 ∈ R3 is given by q/(4πr), where r = (x2 + y 2 + z 2 )1/2 . The corresponding electric field is defined to be minus the gradient of this potential; that is, E=

qr . 4πr3

R Thus Gauss’ law states that the total electric flux ∂Ω E · n dS equals q if 0 ∈ Ω and equals zero, if 0 ∈ / Ω. A continuous charge distribution in Ω described by a charge density ρ is related to E by ρ = div E. By Gauss’ theorem the electric flux Z Z E · n dS = ∂Ω

ρ dV, Ω

which represents the total charge inside Ω. Thus, the relationship div E = ρ may be phrased as follows. The flux out of a surface of an electric field equals the total charge inside the surface.

Exercises  8.3-1.

Use Green’s theorem to show that

(i) The area of the ellipse x2 /a2 + y 2 /b2 = 1 is πab. (ii) The area of the hypocycloid x = a cos3 θ, y = b sin3 θ is 3πa2 /8. (iii) The area of one loop of the four-leaved rose r = 3 sin 2θ is 9π/8.  8.3-2.

Why does Green’s theorem fail in the unit disk for −y

(x2

dx xdy + 2 ? 2 + y ) (x + y 2 )

8.3 The Classical Theorems of Green, Gauss, and Stokes

 8.3-3.

489

For an oriented surface S and a fixed vector a, show that Z Z 2 a · n dS = (a × r) · dS. S

∂S

 8.3-4. Let the components of the vector field X ∈ X(R3 ) be homogeneous of degree one; that is, X satisfies X i (tx, ty, tz) = tX i (x, y, z), i = 1, 2, 3. Show that if curl X = 0, then X = grad f , where f = (xX 1 + yX 2 + zX 3 )/2.  8.3-5.

Let S be the surface of a region Ω in R3 . Show that Z 1 volume(Ω) = r · ndS. 3 S

Give an intuitive argument why this should be so. Hint: Think of cones. Let S be a closed (i.e., compact boundaryless) oriented surface in R3 . R (i) Show in two ways that S (curl X) · n dS = 0.

 8.3-6.

(ii) Let X = X(S) and f ∈ F(S). Make sense of and show that Z

Z (grad f )X · dS = −

S

(f curl X)dS S

where grad, curl, and div are taken in R3 .  8.3-7. Let X and Y be smooth vector fields on an open set D ⊂ R3 , with ∂D smooth, and cl(D) compact. Show that Z Z Z Y · curl X dV = X · curl Y dV + (X × Y) · n dS. D

D

∂D

where n is the outward unit normal to ∂D and dS the induced surface measure on ∂D. Hint: Show that Y · curl X − X · curl Y = div(X × Y).  8.3-8.

If C is a closed curve bounding a surface S show that Z Z Z f (grad g) · ds = (grad f × grad g) · n dS = − g(grad f ) · ds C

S

C

where f and g are C 2 functions.  8.3-9 (A. Lenard). Faraday’s law relates the line integral of the electric field around a loop C to the surface integral of the rate of change of the magnetic field over a surface S with boundary C. Regarding the equation ∇ × E = −∂B/∂t as the basic equation, Faraday’s law is a consequence of Stokes’ theorem, as we have seen in Example 8.3.5B. Suppose we are given electric and magnetic fields in space that satisfy the equation ∇ × E = −∂B/∂t. Suppose C is the boundary of the M¨obius band shown in Figure 7.5.1. Since the M¨obius band cannot be oriented, Stokes’ theorem does not apply. What becomes of Faraday’s law? Resolve the issue in two ways: (i) by using the results of Supplement 8.2A or a direct reformulation of Stokes’ theorem for nonorientable surfaces, and (ii) realizing C as the boundary of an orientable surface. If ∂B/∂t is arbitrary, in general does a current flow around C or not?

490

8.4

8. Integration on Manifolds

Induced Flows on Function Spaces and Ergodicity

This section requires some results from functional analysis. Specifically we shall require a knowledge of Stone’s theorem and self-adjoint operators. The required results may be found in Supplements 8.4A and 8.4B at the end of this section. Flows on manifolds induce flows on tangent bundles, tensor bundles, and spaces of tensor fields by means of push-forward. In this section we shall be concerned mainly with the induced flow on the space of functions. This induced flow is sometimes called the Liouville flow. Let M be a manifold and µ a volume element on M ; that is, (M, µ) is a volume manifold. If Ft is a (volume-preserving) flow on M , then Ft induces a linear one-parameter group (of isometries) on the Hilbert space H = L2 (M, µ) by Ut (f ) = f ◦ F−t . The association of Ut with Ft replaces a nonlinear finite-dimensional problem with a linear infinite-dimensional one. There have been several theorems that relate properties of Ft and Ut . The best known of these is the result of Koopman [1931], which shows that Ut has one as a simple eigenvalue for all t if and only if Ft is ergodic. (If there are no other eigenvalues, then Ft is called weakly mixing.) A few basic results on ergodic theory are given below. We refer the reader to the excellent texts of Halmos [1956], Arnol’d and Avez [1967], and Bowen [1975] for more information. We shall first present a result of Povzner [1966], which relates the completeness of the flow of a divergencefree vector field X to the skew-adjointness of X as an operator. (The hypothesis of divergence free is removed in Exercises 8.4-1–8.4-3.) We begin with a lemma due to Ed Nelson. 8.4.1 Lemma. Let A be an (unbounded ) self-adjoint operator on a complex Hilbert space H. Let D0 ⊂ D(A) (the domain of A) be a dense linear subspace of H and suppose Ut = eitA (the unitary one-parameter group generated by A) leaves D0 invariant. Then A0 := (A restricted to D0 ) is essentially self-adjoint; that is, the closure of A0 is A. Proof. Let A denote the closure of A0 . Since A is closed and extends A0 , A extends A. We need to prove that A extends A. For λ > 0, λ − iA is surjective with a bounded inverse. First of all, we prove that λ − iA0 has dense range. If not, there is a v ∈ H such that hv, λx − iA0 xi = 0 for all x ∈ D0 . In particular, since D0 is Ut -invariant, d hv, Ut xi = hv, iAUt xi = λhv, Ut xi dt so hv, Ut xi = eλt hv, xi. Since D0 is dense, this holds for all x ∈ H. Thus, kUt k = 1 and λ > 0 imply v = 0. Therefore (λ − iA0 )−1 makes sense and (λ − iA)−1 is its closure. It follows from Supplement 8.4A that A is the closure of A0 (see Corollary 8.4.15 and the remarks following it).  8.4.2 Proposition. Let X be a C ∞ divergence-free vector field on (M, µ) with a complete flow Ft . Then iX is an essentially self-adjoint operator on Cc∞ = C ∞ functions with compact support in the complex Hilbert space L2 (M, µ). Proof. Let Ut f = f ◦ F−t be the unitary one-parameter group induced from Ft . A straightforward convergence argument shows that Ut f is continuous in t in L2 (M, µ). In Lemma 8.4.1, choose D0 = C ∞ functions

8.4 Induced Flows on Function Spaces and Ergodicity

491

with compact support. This is clearly invariant under Ut . If f ∈ D0 , then d d = = −df · X, Ut f f ◦ F−t dt dt t=0 t=0 so the generator of Ut is an extension of −X (as a differential operator) on D0 . The corresponding essentially self–adjoint operator is therefore iX.  Now we prove the converse of Proposition 8.4.2. That is, if iX is essentially self-adjoint, then X has an almost everywhere complete flow. This then gives a functional-analytic characterization of completeness. 8.4.3 Theorem. Let M be a manifold with a volume element µ and X be a C ∞ divergence-free vector field on M . Suppose that, as an operator on L2 (M, µ), iX is essentially self-adjoint on the C ∞ functions with compact support. Then, except possibly for a set of points x of measure zero, the flow Ft (x) of X is defined for all t ∈ R. We shall actually prove that, if the defect index of iX is zero in the upper half-plane (i.e., if (iX + i)(Cc∞ ) is dense in L2 ), then the flow is defined, except for a set of measure zero, for all t > 0. Similarly, if the defect index of iX is zero in the lower half-plane, the flow is essentially complete for t < 0. The converses of these more general results can be established along the lines of the proof of Lemma 8.4.1. Proof (E. Nelson—private communication). Suppose that there is a set E of finite positive measure such that if x ∈ E, Ft (x) fails to be defined S for t sufficiently large. Let ET be the set of x ∈ E for which Ft (x) is undefined for t ≥ T . Since E = T ≥1 ET , some ET has positive measure. Replacing E by ET , we may assume that all points of E “move to infinity” in a time ≤ T . If f is any function on M , we adopt the convention that f (Ft (x)) = 0 if Ft (x) is undefined. For any x ∈ M , and t < −T , Ft (x) must be either in the complement of E or undefined; otherwise it would be a point of E that did not move to infinity in time T . Hence χE (Ft (x)) = 0 for t < −T, where χE is the characteristic function of E. We now define a function on M by Z

∞

g(x) =

e−τ χE (Fτ (x))dτ.

−∞

Note that the integral converges because the integrand vanishes for t < −T . In fact, we have Z

∞

0 ≤ g(x) ≤

e−τ dτ = eT .

−T

Moreover, g is in L2 . Indeed, because Ft is measure-preserving, where defined, denoting by k · k2 the L2 norm, we have kχE ◦ Fτ k2 ≤ kχE k2 , so that Z

∞

kgk2 ≤

e−τ kχE ◦ Fτ k2 dτ ≤ kχE k2 eT .

−T

The function g is nonzero because E has positive measure. Fix a point x ∈ M . Then Ft (x) is defined for t sufficiently small. It is easy to see that in this case Fτ (Ft (x)) and Fτ +t (x) are defined or undefined together, and in the former case they are equal. Hence we have χE (Fτ (Ft (x)) = χE (Fτ +t (x)) for t sufficiently small. Therefore, for Z

∞

g(Ft (x)) = −∞

e−τ χE (Fτ +t (x))dτ =

Z

∞

−∞

et−τ χE (Fτ (x))dτ = et g(x).

492

8. Integration on Manifolds

Now if ϕ is C ∞ with compact support, we have Z Z ϕ(Ft (x)) − ϕ(x) g(x)X[ϕ](x)dµ = lim g(x) dµ t→0 t Z g(F−t (x)) − g(x) = lim ϕ(x)dµ t→0 t Z −t Z e −1 = lim g(x)ϕ(x)dµ = − g(x)ϕ(x)dµ. t→0 t These equalities are justified because on the support of ϕ the flow Ft exists for sufficiently small t and is measure-preserving. Thus g is orthogonal to the range of X + 1, and therefore the defect index of iX in the upper half-plane is nonzero. The case of completeness for t < 0 is similar.  Methods of functional analysis applied to L2 (M, µ) can, as we have seen, be used to obtain theorems relevant to flows on M . Related to this is a measure-theoretic analogue of the fact that any automorphism of the algebra F(M ) is induced by a diffeomorphism of M (see Supplement 4.2C). This result, due to Mackey [1962], states that if Ut is a linear isometry on L2 (M, µ), which is multiplicative (i.e., Ut (f g) = (Ut f )(Ut g), where defined), then Ut is induced by some measure preserving flow Ft on M . This may be used to give another proof of Theorem 8.4.3. An important notion for statistical mechanics is that of ergodicity; this is intended to capture the idea that a flow may be random or chaotic. In dealing with the motion of molecules, the founders of statistical mechanics, particularly Boltzmann and Gibbs, made such hypotheses at the outset. One of the earliest precise definitions of randomness of a dynamical system was minimality : the orbit of almost every point is dense. In order to prove useful theorems, von Neumann and Birkhoff in the early 1930s required the strong assumption of ergodicity, defined as follows. 8.4.4 Definition. Let S be a measure space and Ft a (measurable) flow on S. We call Ft ergodic if the only invariant measurable sets are ∅ and all of S. Here, invariant means Ft (A) = A for all t ∈ R and we agree to write A = B if A and B differ by a set of measure zero. (It is not difficult to see that ergodicity implies minimality if we are on a second countable Borel space.) A function f : S → R will be called a constant of the motion if f ◦ Ft = f a.e. (almost everywhere) for each t ∈ R. 8.4.5 Proposition. Proof.

A flow Ft on S is ergodic iff the only constants of the motion are constant a.e.

If Ft is ergodic and f is a constant of the motion, the two sets { x ∈ S | f (x) ≥ a }

and { x ∈ S | f (x) ≤ a } are invariant, so f must be constant a.e. The converse follows by taking f to be a characteristic function.  The first major step in ergodic theory was taken by von Neumann [1932], who proved the mean ergodic theorem which remains as one of the most important basic theorems. The setting is in Hilbert space, but we shall see how it applies to flows of vector fields in Corollary 8.4.7. 8.4.6 Theorem (Mean Ergodic Theorem). Let H be a real or complex Hilbert space and Ut : H → H a strongly continuous one-parameter unitary group (i.e., Ut is unitary for each t, is a flow on H and for each x ∈ H, the map t 7→ Ut x is continuous). Let the closed subspace H0 be defined by H0 = { x ∈ H | Ut x = x for all t ∈ R }

8.4 Induced Flows on Function Spaces and Ergodicity

493

and let P be the orthogonal projection onto H0 . Then for any x ∈ H,

Z t

1

Us x ds − Px lim

= 0. t→∞ t 0 The point av(x) = Px so defined is called the time average of x. Proof (Riesz [1944]).

We must show that

Z t

1

= 0. lim U x ds − Px s

t→∞ t 0

If Px = x, this means x ∈ H0 , so Us (x) = x; the result is clearly true in this case. We can therefore suppose that Px = 0 by considering the decomposition x = Px + (x − Px). Note that { Ut y − y | y ∈ H, t ∈ R }⊥ = H0 where ⊥ denotes the orthogonal complement. This is an easy verification using unitarity of Ut and Ut−1 = U−t . It follows that ker P is the closure of the space spanned by elements of the form Us y − y. Indeed ⊥ ⊥ ker P = H⊥ is the closure of the span of A. Therefore, for 0 , and if A is any set in H, and B = A , then B any  > 0, there exists t1 , . . . , tn and x1 , . . . , xn such that



n X

x − (Utj xj − xj )

< .

j=1 It follows from this, again using unitarity of Ut , that it is enough to prove our assertion for x of the form Uτ y − y. Thus we must establish Z 1 t lim Us (Uτ y − y)ds = 0. t→∞ t 0 For t > τ we may estimate this integral as follows:

Z t



Z Z

1

1 τ

1 t+τ

(Us Uτ y − Us y)ds = − Us (y)ds + Us (y)ds

t

t 0 t 0 0 Z τ Z t+τ 1 1 kykds + kykds ≤ t 0 t 0 2τ kyk = → 0 as t → ∞. t



To apply Theorem 8.4.6 to a measure-preserving flow Ft on S, we consider the unitary one-parameter group Ut (f ) = f ◦ Ft on L2 (S, µ). We only require a minimal amount of continuity on Ft here, namely, we assume that if s → t, Fs (x) ∈ Ft (x) for a.e. x ∈ S. We shall also assume µ(S) < ∞ for convenience. Under these hypotheses, Ut is a strongly continuous unitary one–parameter group. The verification can be done with the aid of the dominated convergence theorem. 8.4.7 Corollary.

In the hypotheses above Ft is ergodic if and only if for each f ∈ L2 (S) its time average Z 1 t av(f )(x) = lim (f ◦ Fs )(x)ds t→±∞ t 0

2 (the limit being in R the L -mean) is constant a.e. In this case the time average av(f ) necessarily equals the space average S f dµ/µ(S) a.e.

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8. Integration on Manifolds

Proof. Ergodicity of Ft is equivalent, by Proposition 8.4.5, to dim H0 = 1, where H0 is the closed subspace of L2 (S) given in Theorem 8.4.6. If dim H0 = 1, Z f dµ P(f ) = µ(S) S so the equality of av(f ) with P(f ) a.e. is a consequence of Theorem 8.4.6. Conversely if any f ∈ L2 (S) has a.e. constant time average av(f ) then taking f to be a constant of motion, it follows that f = av(f ) is constant a.e. Therefore, dim H0 = 1.  Thus, if Ft is ergodic, the time average of a function is constant a.e. and equals its space average. A refinement of this is the individual ergodic theorem of Birkhoff [1931], in which one obtains convergence almost everywhere. Also, if µ(S) = ∞ but f ∈ L1 (S) ∩ L2 (S), one still concludes a.e. convergence of the time average. (If f is only L2 , mean convergence to zero is still assured by Proposition 8.4.5.) Modern work in dynamical systems, following the ideas in §4.3, has shown that for many interesting flows arising in the physical sciences, the motion can be “chaotic” on large regions of phase space without being ergodic. Much current research is focused on trying to prove analogues of the ergodic theorems for such cases. (See, for instance, Guckenheimer and Holmes [1983], Eckmann and Ruelle [1985], and references therein.) A particularly important example of an ergodic flow is the irrational flow on the torus. 8.4.8 Definition. The flow Ft : Tn → Tn given by Ft ([ϕ]) = [ϕ + νt], for ν ∈ Rn is called the quasiperiodic or linear flow on Tn determined by ν. The quasiperiodic flow is called irrational if the components (ν 1 , . . . , ν n ) of ν are linearly independent over Z (or, equivalently, over Q), that is, k · ν = 0 for k ∈ Zn implies k = 0. 8.4.9 Proposition.

The linear flow Ft on Tn determined by ν ∈ Rn is ergodic if and only if it is irrational.

Proof. Assume the flow is irrational and let f ∈ L2 (Tn ) be a constant of the motion. Expand f ([ϕ]) and (f ◦ Ft )([ϕ]) in Fourier series: X X f ([ϕ]) = ak eik·ϕ and (f ◦ Ft )([ϕ]) = bk (t)eik·ϕ k∈Zn

k∈Zn

where the convergence is in L2 and the Fourier coefficients are given by Z ak = eik·ϕ f ([ϕ])dϕ n ZT bk (t) = eik·ϕ f ([ϕ + νt])dϕ Tn Z = eik·(ϕ−νt) f ([ϕ])dϕ = eik·νt ak . Tn

(The measure dϕ is chosen such that the total volume of Tn equals one.) Since f is a constant of the motion, ak = bk (t) for all k ∈ Zn and all t ∈ R which implies that ek·νt = 1 for all k ∈ Zn , t ∈ R. Thus k · ν = 0 which by hypothesis forces k = 0. Consequently all ak = 0 with the exception of a0 and thus f = a0 a.e. Conversely, assume Ft is ergodic and that k · ν = 0 for some k ∈ Zn \{0}. Then the set A = { [ψ] ∈ Tn | k · ψ = 0 } is closed and hence measurable and invariant under Ft . But clearly A 6= ∅ and A 6= Tn which shows that Ft is not ergodic.  8.4.10 Corollary. Let Ft be an irrational flow on Tn determined by ν. Then every trajectory of Ft is uniformly distributed on Tn , that is, for any measurable set A in Tn , lim

t→±∞

measure A(t) = measure A t

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495

where A(t) = { s ∈ [0, t] | Fs ([ψ]) ∈ A } n

and the measure of T is assumed to be equal to one. Proof.

Let χA be the characteristic function of A. Then Z 1 t av(χA )([ψ]) = lim χA (Fs ([ψ]))ds t→±∞ t 0 1 = lim (measure A(t)) t→±∞ t Z =

(χA ([ψ]))dϕ Tn

= measure A by Corollary 8.4.7 and Proposition 8.4.9. 8.4.11 Corollary. irrational.



Every trajectory of a quasiperiodic flow Ft on Tn is dense if and only if the flow is

Proof. By translation of the initial condition it is easily seen that every trajectory is dense on Tn if and only if the trajectory through [0] is dense. Assume first that the flow is irrational. If { Ft ([0]) | t ∈ R } is not dense in Tn then there is an open set U in Tn not containing any point of this trajectory. Thus, in the notation of Corollary 8.4.10, U (t) = ∅. This contradicts 8.4.10 since the measure of U is strictly positive. Conversely, assume that the trajectory through [0] is dense and let f be a continuous constant of the motion for Ft . This implies that f is a constant. Since continuous functions are dense in L2 , this in turn implies that any L2 -constant of the motion is constant a.e. By Proposition 8.4.5, Ft is ergodic and by Proposition 8.4.9, Ft is irrational. 

Supplement 8.4A Unbounded and Self Adjoint Operators.3 In many applications involving differential equations, the operators one meets are not defined on the whole Banach space E and are not continuous. Thus we are led to consider a linear transformation A : DA ⊂ E → E where DA is a linear subspace of E (the domain of A). If DA is dense in E, we say A is densely defined . We speak of A as an operator and this shall mean linear operator unless otherwise specified. Even though A is not usually continuous, it might have the important property of being closed. We say A is closed if its graph ΓA ΓA = { (x, Ax) ∈ E × E | x ∈ DA } is a closed subset of E × E. This is equivalent to (xn ∈ DA , xn → x ∈ E and Axn → y ∈ E) implies (x ∈ DA and Ax = y). An operator A (with domain DA ) is called closable if cl(ΓA ), the closure of the graph of A, is the graph of an operator, say, A. We call A the closure of A. It is easy to see that A is closable iff {(xn ∈ DA , xn → 0 3 This

supplement was written in collaboration with P. Chernoff.

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and Axn → y) implies y = 0}. Clearly A is a closed operator that is an extension of A; that is, DA ⊃ DA and A = A on DA . One writes this as A ⊃ A. The closed graph theorem from §2.2 asserts that an everywhere defined closed operator is bounded. However, if an operator is only densely defined, “closed” is weaker than “bounded.” If A is a closed operator, the map x 7→ (x, Ax) is an isomorphism between DA and the closed subspace ΓA . Hence if we set |||x|||2 = kxk2 + kAxk2 , DA becomes a Banach space. We call the norm ||| · ||| on DA the graph norm. Let A be an operator on a real or complex Hilbert space H with dense domain DA . The adjoint of A is the operator A∗ with domain DA∗ , defined as follows: DA∗ = { y ∈ H | there is a z ∈ H such that hAx, yi = hx, zi for all x ∈ DA } and A∗ : DA∗ → H,

y 7→ z.

∗

From the fact that DA is dense, we see that A is indeed well defined (there is at most one such z for any y ∈ H). It is easy to see that if A ⊃ B then B ∗ ⊃ A∗ . If A is everywhere defined and bounded, it follows from the Riesz representation theorem (Supplement 2.2A) that A∗ is everywhere defined; moreover it is not hard to see that, in this case, kA∗ k = kAk. An operator A is symmetric (Hermitian in the complex case) if A∗ ⊃ A; that is, hAx, yi = hx, Ayi for all x, y ∈ DA . If A∗ = A (this includes the condition DA∗ = DA ), then A is called self-adjoint. An everywhere defined symmetric operator is bounded (from the closed graph theorem) and so is self-adjoint. It is also easy to see that a self-adjoint operator is closed. One must be aware that, for technical reasons, it is the notion of self-adjoint rather than symmetric, which is important in applications. Correspondingly, verifying self-adjointness is often difficult while verifying symmetry is usually trivial. Sometimes it is useful to have another concept at hand, that of essential self-adjointness. First, it is easy to check that any symmetric operator A is closable. The closure A is easily seen to be symmetric. One says that A is essentially self-adjoint when its closure A is self-adjoint. Let A be a self-adjoint operator. A dense subspace C ⊂ H is said to be a core of A if C ⊂ DA and the closure of A restricted to C is again A. Thus if C is a core of A one can recover A just be knowing A on C. We now give a number of propositions concerning the foregoing concepts, which are useful in applications. Most of this is classical work of von Neumann. We begin with the following. 8.4.12 Proposition. Let A be a closed symmetric operator of a complex Hilbert space H. If A is selfadjoint then A + λI is surjective for every complex number λ with Im λ 6= 0 (I is the identity operator ). Conversely, if A is symmetric and A − iI and A + iI are both surjective then A is self-adjoint. Proof.

Let A be self-adjoint and λ = α + iβ, β 6= 0. For x ∈ DA we have k(A + λ)xk2 = k(A + α)xk2 + iβhx, Axi − iβhAx, xi + β 2 kxk2 = k(A + α)xk2 + β 2 kxk2 ≥ β 2 kxk2 ,

where A + λ means A + λI. Thus we have the inequality k(A + λ)xk ≥ | Im λ| kxk

(8.4.1)

Since A is closed, it follows from equation (8.4.1) that the range of A + λ is a closed set for Im λ 6= 0. Indeed, let yn = (A + λ)xn → y. By the inequality (8.4.1), kxn − xm k ≤

kyn − ym k | Im λ|

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497

so xn converges to, say x. Also Axn converges to y − λx; thus x ∈ DA and y − λx = Ax as A is closed. Now suppose y is orthogonal to the range of A + λI. Thus hAx + λx, yi = 0 for all x ∈ DA ,

or hAx, yi = −hx, λyi.

¯ since A = A∗ , y ∈ DA , and Ay = −λy, ¯ we obtain (A + λI)y = 0. By definition, y ∈ DA∗ and A∗ y = −λy; Thus the range of A + λI is all of H. Conversely, suppose A + i and A − i are onto. Let y ∈ DA∗ . Thus for all x ∈ DA , h(A + i)x, yi = hx, (A∗ − i)yi = hx, (A − i)zi for some z ∈ DA since A − i is onto. Thus, h(A + i)x, yi = h(A + i)x, zi and it follows that y = z. This proves that DA∗ ⊂ DA and so DA = DA∗ . The result follows.



If A is self-adjoint then for Im λ 6= 0, λI − A is onto and from equation (8.4.1) is one-to-one. Thus (λI − A)−1 : H → H exists, is bounded, and we have k(λI − A)−1 k ≤

1 | Im λ|

(8.4.2)

This operator (λI −A)−1 is called the resolvent of A. Notice that even though A is an unbounded operator, the resolvent is bounded. The same argument used to prove Proposition 8.4.12 shows the following. 8.4.13 Proposition. are dense.

A symmetric operator A is essentially self-adjoint iff the ranges of A + iI and A − iI

If A is a (closed) symmetric operator then the ranges of A + iI and A − iI are (closed) subspaces. The dimensions of their orthogonal complements are called the deficiency indices of A. Thus, Propositions (8.4.12) and (8.4.13) can be restated as: a closed symmetric operator (resp., a symmetric operator ) is selfadjoint (resp., essentially self-adjoint) iff it has deficiency indices (0, 0). If A is a closed symmetric operator then from equation (8.4.1), A + iI is one-to-one and we can consider the inverse (A + iI)−1 , defined on the range of A + iI. One calls (A − iI)(A + iI)−1 the Cayley transform of A. It is always isometric, as is easy to check. Thus A is self–adjoint iff its Cayley transform is unitary. Let us return to the graph of an operator A for a moment. The adjoint can be described entirely in terms of its graph and this is often convenient. Define an isometry J : H ⊕ H → H ⊕ H by J(x, y) = (−y, x); note that J 2 = −I. 8.4.14 Proposition. Let A be densely defined. Then (ΓA )⊥ = J(ΓA∗ ) and −ΓA∗ = J(ΓA )⊥ . In particular, A∗ is closed, and if A is closed, then H ⊕ H = ΓA ⊕ J(ΓA∗ ), where H ⊕ H carries the usual inner product: h(x1 , x2 ), (y1 , y2 )i = hx1 , y1 i + hx2 , y2 i . Proof.

Let (z, y) ∈ J(ΓA∗ ), so y ∈ DA∗ and z = −A∗ y. Let x ∈ DA . We have h(x, Ax), (−A∗ y, y)i = hx, −A∗ yi + hAx, yi = 0,

and so J(ΓA∗ ) ⊂ Γ⊥ A. Conversely if (z, y) ∈ (ΓA )⊥ , then hx, zi + hAx, yi = 0 for all x ∈ DA . Thus by definition, y ∈ DA∗ and z = −A∗ y. This proves the opposite inclusion. 

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Thus if A is a closed operator, the statement H ⊕ H = ΓA ⊕ J(ΓA∗ ) means that given e, f ∈ H, the equations x − A∗ y = e and Ax + y = f have exactly one solution (x, y). If A is densely defined and symmetric, then its closure A satisfies A ⊂ A∗ since A∗ is closed. There are other important consequences of Proposition 8.4.14 as well. 8.4.15 Corollary.

For A densely defined and closeable, we have

∗∗

(i) A = A , and (ii) A∗ = A∗ . Proof.

(i) Note that ΓA∗∗ = −J{(ΓA∗ )⊥ } = −(J(ΓA∗ ))⊥

since J is an isometry. But ⊥ ⊥⊥ −(J(ΓA∗ ))⊥ = −(J 2 Γ⊥ A ) = ΓA = cl(ΓA ) = ΓA . ⊥ (ii) follows since Γ⊥ A = cl(ΓA ).



Suppose A : DA ⊂ H → H is one-to-one. Then we get an operator A−1 defined on the range of A. In terms of graphs: ΓA−1 = K(ΓA ), where K(x, y) = (y, x); note that K 2 = I, K is an isometry and KJ = −JK. It follows for example that if A is self-adjoint, so is A−1 , since ⊥ ⊥ Γ(A−1 )∗ = −J(Γ⊥ A−1 ) = −J(KΓA ) = KJΓA = KΓ(A∗ ) = ΓA∗−1 .

Next we consider possible self-adjoint extensions of a symmetric operator. 8.4.16 Proposition. are equivalent:

Let A be a symmetric densely defined operator on H and A its closure. The following

(i) A is essentially self-adjoint. (ii) A∗ is self-adjoint. (iii) A∗∗ ⊃ A∗ . (iv) A has exactly one self-adjoint extension. (v) A = A∗ . Proof. By definition, (i) means A∗ = A. But we know A∗ = A∗ and A = A∗∗ by Corollary 8.4.15. Thus (i), (ii), (v) are equivalent. These imply (iii). Also (iii) implies (ii) since A ⊂ A ⊂ A∗ ⊂ A∗∗ = A and so A∗ = A∗∗ . To prove (iv) is implied let Y be any self-adjoint extension of A. Since Y is closed, Y ⊃ A. But A = A∗ so Y extends the self-adjoint operator A∗ ; that is, Y ⊃ A. Taking adjoints, A∗ = A ⊃ Y ∗ = Y so Y = A. To prove that (iv) implies the others is a bit more complicated. We shall in fact give a more general result in Proposition 8.4.18 below. First we need some notation. Let D+ = range(A + iI)⊥ ⊂ H and D− = range(A − iI)⊥ ⊂ H called the positive and negative defect spaces. Using the argument in Proposition 8.4.12 it is easy to check that D+ = { x ∈ DA∗ | A∗ x = ix } and D− = { x ∈ DA∗ | A∗ x = −ix }. 

8.4 Induced Flows on Function Spaces and Ergodicity

8.4.17 Lemma.

499

Using the graph norm on DA∗ , we have the orthogonal direct sum DA∗ = DA ⊕ D+ ⊕ D− .

Proof. Since D+ , D− are closed in H they are closed in DA∗ . Also DA ⊂ DA∗ is closed since A∗ is an extension of A and hence of A. It is easy to see that the indicated spaces are orthogonal. For example let x ∈ DA and y ∈ D− . Then using the inner product hhx, yii = hx, yi + hA∗ x, A∗ yi gives hhx, yii = hx, yi + hA∗ x, −iyi = hx, yi − ihA∗ x, yi. Since x ∈ DA = DA∗ , by Proposition 8.4.16(v), we get hhx, yii = hx, yi − ihx, A∗ yi = hx, yi − hx, yi = 0. To see that DA∗ = DA ⊕ D+ ⊕ D− it suffices to show that the orthogonal complement of DA ⊕ D+ ⊕ D− is zero. Let u ∈ (DA ⊕ D+ ⊕ D− )⊥ , so hhu, xii = hhu, yii = hhu, zii = 0 for all x ∈ DA , y ∈ D+ , z ∈ D− . From hhu, xii = 0 we get hu, xi + hA∗ u, A∗ xi = 0 that is, or A∗ u ∈ DA∗ and A∗ A∗ u = −u. It follows that (I − iA∗ )u ∈ D+ . But from hhu, yii = 0 we have h(I − iA∗ )u, yi = 0 and so (I − iA∗ )u = 0. Hence u ∈ D− . Taking z = u gives u = 0.  8.4.18 Proposition. The self-adjoint extensions of a symmetric densely defined operator A (if any) are obtained as follows. Let T : D+ → D− be an isometry mapping D+ onto D− and let ΓT ⊂ D+ ⊕ D− be its graph. Then the restriction of A∗ to DA ⊕ ΓT is a self-adjoint extension of A. Thus, A has self-adjoint extensions iff its defect indices (dim D+ , dim D− ) are equal and these extensions are in one-to-one correspondence with all isometries of D+ onto D− . Assuming this result for a moment, we give the following. Completion of Proof of Proposition 8.4.16. If there is only one self-adjoint extension it follows from proposition 8.4.18 that D+ = D− = {0} so by proposition 8.4.13, A is essentially self-adjoint.  Proof of Proposition 8.4.18. Let B be a self-adjoint extension of A. Then A∗ = A∗ ⊃ B so B is the restriction of A∗ to some subspace containing DA . We want to show that these subspaces are of the form DA ⊕ ΓT as stated. Suppose first that T : D+ → D− is an isometry onto and let A be the restriction of A∗ to DA ⊕ ΓT . First of all, one proves that A is symmetric: that is, for u, x ∈ DA and v, y ∈ D+ that hAx + A∗ y + A∗ T y, u + v + T vi = hx + y + T y, Au + A∗ v + A∗ T vi. This is a straightforward computation using the definitions. To show that A is self-adjoint, we show that DA∗ ⊂ DA . If this does not hold there exists a nonzero z ∈ DA∗ such that either A∗ z = iz or A∗ z = −iz. This follows from Lemma 8.4.17 applied to the operator A. (Observe that A is a closed operator—this easily follows.) Now A ⊃ A so A∗ ⊃ A∗ . Thus z ∈ D+ or z ∈ D− . Suppose z ∈ D+ . Then z + T z ∈ DA so as hhDA , zii = 0, where hh , ii denotes the inner product relative to A, 0 = hhz + T z, zii = hhz, zii + hhT z, zii = 2hz, zi,

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since T z ∈ D− . Hence z = 0. In a similar way one sees that if z ∈ D− then z = 0. Hence A is self-adjoint. We will leave the details of the converse to the reader (they are similar to the foregoing). The idea is this: if A is restriction of A∗ to a subspace DA ⊕ V for V ⊂ D+ ⊕ D− and A is symmetric, then V is the graph of a map T : W ⊂ D+ → D− and hT u, T vi = hu, vi, for a subspace W ⊂ D+ . Then self-adjointness of A implies that in fact W = D+ and T is onto.  A convenient test for establishing the equality of the deficiency indices is to show that T commutes with a conjugation U ; that is, an antilinear isometry U : H → H satisfying U 2 = I; antilinear means that U (αx) = α ¯U x for complex scalars α and U (x + y) = U x + U y for x, y ∈ H. It is easy to see that U is the isometry required from D+ to D− (use D+ = range(A + iI)⊥ ). As a corollary, we obtain an important classical result of von Neumann: Let H be L2 of a measure space and let A be a (closed ) symmetric operator that is real in the sense that it commutes with complex conjugation. Then A admits self-adjoint extensions. (Another sufficient condition of a different nature, due to Friedrichs, is given below.) This result applies to many quantum mechanical operators. However, one is also interested in essential self-adjointness, so that the self-adjoint extension will be unique. Methods for proving this for specific operators in quantum mechanics are given in Kato [1951, 1976] and Reed and Simon [1974]. For corresponding questions in elasticity, see Marsden and Hughes [1983]. We now give some additional results that illustrate methods for handling self-adjoint operators. 8.4.19 Proposition. Let A be a self-adjoint and B a bounded self-adjoint operator. Then A + B (with domain DA ) is self-adjoint. If A is essentially self-adjoint on DA then so is A + B. Proof.

A + B is certainly symmetric on DA . Let y ∈ D(A+B)∗ so that for all x ∈ DA , h(A + B)x, yi = hx, (A + B)∗ yi.

The left side is hAx, yi + hBx, yi = hAx, yi + hx, Byi since B is everywhere defined. Thus hAx, yi = hx, (A + B)∗ y − Byi. Hence y ∈ DA∗ = DA and

Ay = A∗ y = (A + B)∗ y − By.

Hence y ∈ DA+B = DA . Let A be the closure of A. For the second part, it suffices to show that the closure of A + B equals A + B. But if x ∈ DA there is a sequence xn ∈ DA such that xn → x, and Axn → Ax. Then Bxn → Bx as B is bounded so x belongs to the domain of the closure of A + B.  In general, the sum of two self-adjoint operators need not be self-adjoint. (See Nelson [1959] and Chernoff [1974] for this and related examples.) 8.4.20 Proposition. Proof.

Let A be a symmetric operator. If the range of A is all of H then A is self-adjoint.

We first observe that A is one-to-one. Indeed let Ax = 0. Then for any y ∈ DA , 0 = hAx, yi = hx, Ayi.

But A is onto and so x = 0. Thus A admits an everywhere defined inverse A−1 , which is therefore self-adjoint. Hence A is self-adjoint (we proved earlier than the inverse of a self-adjoint operator is self-adjoint). 

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501

We shall use these results to prove a theorem that typifies the kind of techniques one uses. 8.4.21 Proposition. Let A be a symmetric operator on H and suppose A ≤ 0; that is hAx, xi ≤ 0 for x ∈ DA . Suppose I − A has dense range. Then A is essentially self-adjoint. Proof.

Note that h(I − A)u, ui = hu, ui − hAu, ui ≥ kuk2

and so by the Schwarz inequality we have k(I − A)uk ≥ kuk. It follows that the closure of I − A which equals I − A has closed range, which by hypothesis must be all of H. By Proposition 8.4.19 I − A is self-adjoint, so by Proposition 8.4.20 A is self-adjoint.  8.4.22 Corollary.

If A is self-adjoint and A ≤ 0, then for any λ > 0, λ − A is onto, (λ − A)−1 exists and k(λ − A)−1 k ≤

Proof.

1 . λ

(8.4.3)

As before, we have h(λ − A)u, ui ≥ λkuk2 ,

which yields k(λ − A)uk ≥ λkuk. As A is closed, this implies that the range of λ − A is closed. If we can show it is dense, the result will follow. Suppose y is orthogonal to the range h(λ − A)u, yi = 0 for all u ∈ DA . This means that (λ − A)∗ y = 0, or since A is self-adjoint, y ∈ DA . Making the choice u = y gives 0 = h(λ − A)y, yi ≥ λkyk2 , so y = 0.



Note that an operator A has dense range iff A∗ is one-to-one; that is, A∗ w = 0 implies w = 0. For a given symmetric operator A, we considered the general problem of self-adjoint extensions of A and classified these in terms of the defect spaces. Now, under different hypotheses, we construct a special self-adjoint extension (even though A need not be essentially self-adjoint). This result is useful in many applications, including quantum mechanics. A symmetric operator A on H is called lower semi-bounded if there is a constant c ∈ R such that hAx, xi ≥ ckxk2 for all x ∈ DA . Upper semi-bounded is defined similarly. If A is either upper or lower semi-bounded then A is called semi-bounded. Observe that if A is positive or negative then A is semibounded. As an example, let A = −∇2 +V where ∇2 is the Laplacian and let V be a real valued continuous function and bounded below, say V (x) ≥ α. Let H = L2 (Rn , C) and DA the C ∞ functions with compact support. Then −∇2 is positive, so hAf, f i = h−∇2 f, f i + hV f, f i ≥ αhf, f i, and thus A is semi-bounded. We already know that this operator is real so has self-adjoint extensions by von Neumann’s theorem. However, the self-adjoint extension constructed below (called the Friedrichs extension) is “natural.” Thus the actual construction is as important as the statement. 8.4.23 Theorem.

A semi-bounded symmetric (densely defined ) operator admits a self-adjoint extension.

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Proof. After multiplying by −1 if necessary and replacing A by A+(1−α)I we can suppose hAx, xi ≥ kxk2 . Consider the inner product on DA given by hhx, yii = hAx, yi. (Using symmetry of A and the preceding inequality one easily checks that this is an inner product.) Let H1 be the completion of DA in this inner product. Since the H1 -norm is stronger than the H-norm, we have H1 ⊂ H (i.e., the injection DA ⊂ H extends uniquely to the completion). Now let H−1 be the dual of H1 . We have an injection of H into H−1 defined as follows: if y is fixed and x 7→ hx, yi is a linear functional on H, it is also continuous on H1 since |hx, yi| ≤ kxk kyk ≤ |||x||| ||y||, where ||| · ||| is the norm of H1 . Thus H1 ⊂ H ⊂ H−1 . Now the inner product on H1 defines an isomorphism B : H1 → H−1 . Let C be the operator with domain DC = { x ∈ H1 | B(x) ∈ H }, and Cx = Bx for x ∈ DC . Thus C is an extension of A. This will be the extension we sought. We shall prove that C is self-adjoint. By definition, C is surjective; in fact C : DC → H is a linear isomorphism. Thus by Proposition 8.4.20 it suffices to show that C is symmetric. Indeed for x, y ∈ DC we have, by definition, hCx, yi = hhx, yii = hhy, xii = hCy, xi = hx, Cyi.



The self-adjoint extension C can be alternatively described as follows. Let H1 be as before and let C be the restriction of A∗ to DA∗ ∩ H1 . We leave the verification as an exercise.

Supplement 8.4B Stone’s Theorem4 Here we give a self-contained proof of Stone’s theorem for unbounded self-adjoint operators A on a complex Hilbert space H. This guarantees that the one-parameter group eitA of unitary operators exists. In fact, there is a one-to-one correspondence between self-adjoint operators and continuous one-parameter unitary groups. A continuous one-parameter unitary group is a homomorphism t 7→ Ut from R to the group of unitary operators on H, such that for each x ∈ H the map t 7→ Ut x is continuous. The infinitesimal generator A of Ut is defined by Uh (x) − x d iAx = Ut x = lim , h→0 dt h t=0 its domain D consisting of those x for which the indicated limit exists. We insert the factor i for convenience; iA is often called the generator . 8.4.24 Theorem (Stone’s Theorem). Let Ut be a continuous one-parameter unitary group. Then the generator A of Ut is self-adjoint. (In particular, by Supplement 8.4A, it is closed and densely defined.) Conversely, let A be a given self-adjoint operator. Then there exists a unique one-parameter unitary group Ut whose generator is A. 4 This

supplement was written in collaboration with P. Chernoff.

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503

Before we begin the proof, let us note that if A is a bounded self-adjoint operator then one can form the series 1 1 Ut = eitA = I + itA + (itA)2 + (itA)3 + . . . 2! 3! which converges in the operator norm. It is straightforward to verify that Ut is a continuous one-parameter unitary group and that A is its generator. Because of this, one often writes eitA for the unitary group whose generator is A even if A is unbounded. (In the context of the so-called “operational calculus” for self-adjoint operators, one can show that eitA really is the result of applying the function eit(·) to A; however, we shall not go into these matters here.) Proof of Stone’s Theorem (first half). Let Ut be a given continuous unitary group. In a series of lemmas, we shall show that the generator A of Ut is self-adjoint.  8.4.25 Lemma. x ∈ D. Proof.

The domain D of A is invariant under each Ut , and moreover AUt x = Ut Ax for each

Suppose x ∈ D. Then 1 (Uh Ut x − Ut x) = Ut h



 1 (Uh x − h) . h

which converges to Ut (iAx) = iUt Ax as h → 0. The lemma follows by the definition of A. 8.4.26 Corollary. Proof.



The operator A is closed.

If x ∈ D then, by Lemma 8.4.25 d Ut x = iAUt x = iUt Ax. dt

Hence, t

Z Ut x = x + i

Uτ Ax dτ.

(8.4.4)

0

Now suppose that xn ∈ D, xn → x, and Axn → y. Then we have, by equation (8.4.4),   Z t Ut x = lim Ut xn = lim xn + i Uτ Axn dτ . n→∞

n→∞

0

Thus, Z Ut x = x + i

t

Uτ y dτ.

(8.4.5)

0

(Here we have taken the limit under the integral sign because the convergence is uniform; indeed kUτ Axn − Uτ yk = kAxn − yk → 0 independent of τ ∈ [0, t].) Then, by equation (8.4.5), d Ut x = iy. dt t=0 Hence x ∈ D and y = Ax. Thus A is closed. 8.4.27 Lemma.

A is densely defined.



504

8. Integration on Manifolds

Proof.

Let x ∈ H, and let ϕ be a C ∞ function with compact support on R. Define Z

∞

xϕ =

ϕ(t)Ut x dt. −∞

We shall show that xϕ is in D, and that x = limn→∞ xϕn for a suitable sequence {ϕn }. To take the latter point first, let ϕn (t) be nonnegative, zero outside the interval [0, 1/n], and such that ϕn (t) has integral 1. By continuity, if  > 0 is given, one can find N so large that kUt x − xk <  if |t| < 1/N . Suppose that n > N . Then Z 1/n Z ∞ ϕn (t)(Ut x − x)dt ϕn (t)(Ut x − x)dt = kxϕn − xk = 0 −∞ Z 1/n Z 1/n ≤ ϕn (t)kUt x − xkdt ≤  ϕn (t)dt = . 0

0

Finally, we show that xϕ ∈ D; moreover, we shall show that iAxϕ = −xϕ . Indeed, Z

t

−

t

Z

Z

Uτ xϕ0 dτ = 0

∞

ϕ0 (σ)Uσ dσ

Uτ dτ −∞

0

Z

∞

=−

0

Z

t

dσ · ϕ (σ) · −∞ Z ∞

=− −∞

Uτ +σ x dτ 0

dσ · ϕ0 (σ) ·

Z

σ+t

Uτ x dτ. σ

Integrating by parts and using the fact that ϕ has compact support, we get Z −

t

Z

∞

Z

−∞

0

∞

(Uσ+t x − Uσ x)ϕ(σ)dσ = (Ut − I)

Uτ xϕ0 dτ =

Uσ xϕ (σ)dσ. −∞

That is, Z −

t

Uτ xϕ0 dτ = Ut xϕ − xϕ . 0

from which the assertion follows.



Thus far we have made no significant use of the unitarity of Ut . We shall now do so. 8.4.28 Lemma. Proof.

A is symmetric.

Take x, y ∈ D. Then we have 1 d 1 d hUt x, yi = hx, Ut∗ yi i dt i dt t=0 t=0 1 d 1 d = hx, U−t yi =− hx, Ut yi i dt i dt t=0 t=0 1 = − hx, iAyi = hx, Ayi. i

hAx, yi =



8.4 Induced Flows on Function Spaces and Ergodicity

505

To complete the proof that A is self-adjoint, let y ∈ D∗ and x ∈ D. By Lemmas 8.4.25, and 8.4.28,  Z −t  hUt y, xi = hy, U−t xi = hy, xi + y, i Uτ Ax dτ 0 −t

Z

Z

−t

hy, AUτ xidτ

hy, Uτ Axidτ = hy, xi − i

= hy, xi − i

0

0

Z

−t

Z

∗

= hy, xi − i hU−τ A y, xidτ = hy, xi + i 0   Z t ∗ Uτ A y dτ, x . = y+i

−t

hUτ A∗ y, xidτ

0

0

Because D is dense, it follows that t

Z

Uτ A∗ y dτ.

Ut y = y + i 0

Hence, differentiating, we see that y ∈ D and A∗ y = Ay. Thus A = A∗ . Proof of Stone’s theorem (second half). We are now given a self-adjoint operator A. We shall construct a continuous unitary group Ut whose generator is A. 8.4.29 Lemma.

If λ > 0, then I + λA2 : DA2 → H is bijective, (I + λA2 )−1 : H → DA2

is bounded by 1, and DA2 , the domain of A2 , is dense. √ Proof. If A is self-adjoint, so is λA. It is therefore enough to establish the lemma for λ = 1. First we establish surjectivity. By Proposition 8.4.14 and Lemma 8.4.26, if z ∈ H is given there exists a unique solution (x, y) to the equations x − Ay = 0, Ax + y = z. From the first equation, x = Ay. The second equation then yields A2 y + y = z, so I + A2 is surjective. For x ∈ DA2 , note that h(I + A2 )x, xi ≥ kxk2 ,

so

k(I + A2 )xk ≥ kxk.

Thus I + A2 is one-to-one and k(I + A2 )−1 k ≤ 1. Now suppose that u is orthogonal to DA2 . We can find a v such that u = v + A2 v. Then 0 = hu, vi = hv + A2 v, vi = kvk2 + kAvk2 , whence v = 0 and therefore u = 0. Consequently DA2 is dense in H.



For λ > 0, define an operator Aλ by Aλ = A(I + λA2 )−1 . Note that Aλ is defined on all of H because if x ∈ H then (I + λA2 )−1 x ∈ DA2 ⊂ D, so A(I + λA2 )−1 x makes sense. 8.4.30 Lemma. Proof.

Aλ is a bounded self-adjoint operator. Also Aλ and Aµ commute for all λ, µ > 0.

Pick x ∈ H. Then by Lemma 8.4.29 λkAλ xk2 = hλA(I + λA2 )−1 x, A(I + λA2 )−1 xi = hλA2 (I + λA2 )−1 x, (I + λA2 )−1 xi ≤ h(I + λA2 )(I + λA2 )−1 x, (I + λA2 )−1 xi ≤ k(I + λA2 )−1 xk2 ≤ kxk2 ,

506

8. Integration on Manifolds

√ so kAλ k ≤ 1/ λ, and thus Aλ is bounded. We now show that Aλ is self-adjoint. First we shall show that if x ∈ D, then Aλ x = (I + λA2 )−1 Ax. Indeed, if x ∈ D we have Aλ x ∈ DA2 by Lemma 8.4.29, so (I + λA2 )Aλ x = (I + λA2 )A(I + λA2 )−1 x = A(I + λA2 )(I + λA2 )−1 x = Ax. Now suppose x ∈ D and y is arbitrary. Then hAλ x, yi = h(I + λA2 )−1 Ax, yi = h(I + λA2 )−1 Ax, (I + λA2 )(I + λA2 )−1 yi = hAx, (I + λA2 )−1 yi = hx, Aλ yi. Because D is dense and Aλ bounded, this relation must hold for all x ∈ H. Hence Aλ is self-adjoint. The proof that Aλ Aµ = Aµ Aλ is a calculation that we leave to the reader.  Since Aλ is bounded, we can form the continuous one-parameter unitary groups Utλ = eitAλ , λ > 0 using power series or the results of §4.1. Since Aλ and Aµ commute, it follows that Usλ and Utµ commute for every s and t. 8.4.31 Lemma. Proof.

If x ∈ D then limλ→0 Aλ x = Ax.

If x ∈ D we have Aλ x − Ax = (I + λA2 )−1 Ax − Ax = −λA2 (I + λA2 )−1 Ax.

It is therefore enough to show that for every y ∈ H, λA2 (I + λA2 )−1 y → 0. From the inequality k(I + λA2 )yk2 ≥ kλA2 yk2 , valid for λ ≥ 0, we see that kλA2 (I + λA2 )−1 k ≤ 1. Thus it is even enough to show the preceding equality for all y in some dense subspace of H. Suppose y ∈ DA2 , which is dense by Lemma 8.4.29. Then kλA2 (I + λA2 )−1 yk = λk(I + λA2 )−1 A2 yk ≤ λkA2 yk, which indeed goes to zero with λ.



8.4.32 Lemma. For each x ∈ H, limλ→0 Utλ x exists. If we call the limit Ut x, then {Ut } is a continuous one-parameter unitary group. Proof.

We have Z

t

d µ (Uτλ Ut−τ )x dτ dτ 0 Z t   µ µ = iAλ Uτλ Ut−τ x − Uτλ iAu Ut−τ x dτ 0 Z t µ =i Uτλ Ut−τ (Aλ x − Aµ x)dτ,

Utλ x − Utµ x =

0

whence kUtλ x − Utµ xk ≤ |t| kAλ x − Aµ xk.

(8.4.6)

8.4 Induced Flows on Function Spaces and Ergodicity

507

Now suppose that x ∈ D. Then by Lemma 8.4.31, Aλ x → Ax, so that kAλ x − Aµ xk → 0

λ, µ → 0.

as

Because of equation (8.4.6) it follows that {Utλ ξ}λ>0 is uniformly Cauchy as λ → 0 on every compact tinterval. It follows that limλ→0 Utλ x = Ut x exists and is a continuous function of t. Moreover, since D is dense and all of the Utλ have norm 1, an easy approximation argument shows that the preceding conclusion holds even if x ∈ / D. It is obvious that each Ut is a linear operator. Furthermore, hUt x, Ut yi = lim hUtλ x, Utλ yi = lim hx, yi = hx, yi λ→0

λ→0

so Ut is isometric. Trivially, U0 = I. Finally, λ hUs Ut , x, yi = lim hUsλ Ut , x, yi = lim hUt x, U−s yi λ→0

λ→0

λ λ = lim hUtλ x, U−s yi = lim hUs+t x, yi = hUs+t x, yi, λ→0

λ→0

so Us Ut = Us+t . Thus, Us has an inverse, namely U−s , and so Us is unitary. 8.4.33 Lemma.

If x ∈ D, then lim

t→0

Proof.



Ut x − x = iAx. t

We have Utλ x

t

Z

Uτλ Aλ x dτ.

−x=i

(8.4.7)

0

Now Uτλ Aλ x − Uτ Ax = Uτλ (Aλ x − Ax) + Uτλ Ax − Uτ Ax → 0 uniformly for τ ∈ [0, t] as λ → 0. Thus letting λ → 0 in this equation, we get Z t Ut x − x = i Uτ Ax dτ

(8.4.8)

0

for all x ∈ D. The lemma follows directly from equation (8.4.8). 8.4.34 Lemma.



If lim

t→0

Ut x − x = iw t

exists, then x ∈ D. Proof. 8.4.33,

It suffices to show that x ∈ D∗ , the domain of A∗ , since D = D∗ . Let y ∈ D∗ . Then by Lemma     U−t y − y Ut x − x hx, iAyi = lim x, = − lim , y = −hiw, yi. t→0 t→0 −t t

Therefore, hx, Ayi = hw, yi. Thus x ∈ D∗ and so as A is self-adjoint, x ∈ D.



Let us finally prove uniqueness. Let c(t) be a differentiable curve in H such that c(t) ∈ D and c0 (t) = iA(c(t)). We claim that c(t) = Ut c(0). Indeed consider, h(t) = U−t c(t). Then kh(t + τ ) − h(t)k = kU−t−τ c(t + τ ) − U−t−τ Ut c(t)k = kc(t + τ ) − Uτ c(t)k = k(c(t + τ ) − c(t)) − (Uτ c(t) − c(t))k.

508

8. Integration on Manifolds

Hence

h(t + τ ) − h(t) →0 τ as τ → 0, so h is constant. But h(t) = h(0) means c(t) = Ut c(0).



From the proof of Stone’s theorem, one can deduce the following Laplace transform expression for the resolvent, which we give for the sake of completeness. 8.4.35 Corollary.

Let Re λ > 0. Then for all x ∈ H, Z ∞ e−λt Ut xdt. (λ − iA)−1 x = 0

Proof. The foregoing is formally an identity if one thinks of Ut as eitA . Indeed, if A is bounded then it follows just by manipulation of the power series: One has e−λt eitA = e−t(λ−iA) , as one can see by expanding both sides. Next note that Z R h i e−t(λ−iA) xdt = (λ − iA)−1 x − e−R(λ−iA) x , 0

as is seen by integrating the series term by term. Letting R → ∞, one gets the result. Now for arbitrary A we know that Ut x = limµ→0 Utµ x, uniformly on bounded intervals. It follows that Z ∞ Z ∞ −λt e Ut x dt = lim e−λt Utµ x dt = lim (λ − iAµ )−1 x. 0

µ→0

0

µ→0

It remains to show that this limit is (λ − iA)−1 x. Now (λ − iA)−1 x − (λ − iAµ )−1 x = (λ − iAµ )−1 [(λ − iAµ )(λ − iA)−1 x − x]. But (λ − iA)−1 x ∈ D (see Proposition 8.4.12), so by Lemma 8.4.31, (λ − iAµ )(λ − iA)−1 x → (λ − iA)(λ − iA)−1 x = x as µ → 0. Because k(λ − iAµ )−1 k ≤ | Re λ|−1 it follows that k(λ − iA)−1 x − (λ − iAµ )−1 xk → 0.



In closing, we mention that many of the results proved have generalizations to continuous one-parameter groups or semi-groups of linear operators in Banach spaces (or on locally convex spaces). The central result, due to Hille and Yosida, characterizes generators of semi-groups. Our proof of Stone’s theorem is based on methods that can be used in the more general context. Expositions of this more general context are found in, for example, Kato [1976] and Marsden and Hughes [1983, Chapter 6].

Exercises (Exercises 8.4-1–8.4-3 form a unit.)  8.4-1. Given a manifold M , show that the space of half-densities on M carries a natural inner product. Let its completion be denoted H(M ), which is called the intrinsic Hilbert space of M . If µ is a density √ on M , define a bijection of L2 (M, µ) with H(M ) by f 7→ f µ. Show that it is an isometry.

8.5 Introduction to Hodge–deRham Theory

509

 8.4-2. If Ft is the (local) flow of a smooth vector field X, show that Ft induces a flow of isometries on H(M ). (Make no assumption that X is divergence-free.) Show that the generator iX 0 = £X of the induced flow on H(M ) is   1 √ √ µ iX 0 (f µ) = X[f ] + (divµ X)f 2 and check directly that X 0 is a symmetric operator on the space of half-densities with compact support.  8.4-3.

Prove that Ft is complete a.e. if and only if X 0 is essentially self-adjoint.

 8.4-4.

Consider the flow in R2 associated with a reflecting particle: for t > 0, set Ft (q, p) = q + tp

if q > 0, q + tp > 0

and Ft (q, p) = −q − tp

if q > 0, q + tp < 0

and set Ft (−q, p) = −Ft (q, p)

and F−t = Ft−1 .

What is the generator of the induced unitary flow? Is it essentially self-adjoint on the C ∞ functions with compact support away from the line q = 0? Vk 2 2  8.4-5. Let M be R an oriented Riemannian manifold and L ( (M )) the space of L k-forms with inner product hα, βi = α∧∗β. If X is a Killing field on M with a complete flow Ft , show that i£X is a self-adjoint Vk operator on L2 ( (M )).

8.5

Introduction to Hodge–deRham Theory

Recall that a k-form α is called closed if dα = 0 and exact if α = dβ for some k − 1 form β. Since d2 = 0, every exact form is closed, but the converse need not hold. Let H k (M ) =

ker dk , range dk−1

(where dk denotes the exterior derivative on k-forms), and call it the k-th deRham cohomology group of M . (The group structure here is that if a real vector space.) The celebrated deRham theorem states that for a finite-dimensional compact manifold, these groups are isomorphic to the singular cohomology groups (with real coefficients) defined in algebraic topology; the isomorphism is given by integration. For proofs, see Singer and Thorpe [1976] or Warner [1983]. The original books of Hodge [1952] and deRham [1955] (translated as deRham [1984]) remain excellent sources of information as well. A special but important case of the deRham theorem is proved in Supplement 8.5B. The scope of this section is to informally discuss the Hodge decomposition theory based on differential operators and to explain how it is related to the deRham cohomology groups. In addition, some topological applications of the theory are given, such as the Brouwer fixed-point theorem, and the degree of a map is defined. In the sequel, M will denote a compact oriented Riemannian manifold, and δ the codifferential operator. At first we assume M has no boundary. Later we will discuss the case in which M has a boundary. 8.5.1 Definition.

The Laplace–deRham operator ∆ : Ωk (M ) → Ωk (M )

is defined by ∆ = dδ + δd.

510

8. Integration on Manifolds

A form for which ∆α = 0 is called harmonic. Let  Hk = α ∈ Ωk (M ) | ∆α = 0 denote the vector space of harmonic k-forms. If f ∈ H0 (M ), then ∆f = dδf + δdf = δdf = − div grad f, 2

2

so ∆f = −∇ f , where ∇ is the Laplace–Beltrami operator. This minus sign can be a source of confusion and one has to be careful. Recall that the L2 -inner product in Ωk (M ) is defined by Z hα, βi = α ∧ ∗β M

and that d and δ are adjoints with respect to this inner product. That is, hdα, βi = hα, δβi for all α ∈ Ωk−1 (M ), β ∈ Ωk (M ). Thus it follows that for α, β ∈ Ωk (M ), we have h∆α, βi = hdδα, βi + hδdα, βi = hδα, δβi + hdα, dβi = hα, dδβi + hα, δdβi = hα, ∆βi, and thus ∆ is symmetric. This computation also shows that h∆α, αi ≥ 0 for all α ∈ Ωk (M ). 8.5.2 Proposition. Let M be a compact boundaryless oriented Riemannian manifold and α ∈ Ωk (M ). Then ∆α = 0 iff δα = 0 and dα = 0. Proof.

It is obvious from the expression ∆α = dδα + δdα

that if dα = 0 and δα = 0, then ∆α = 0. Conversely, the previous computation shows that 0 = h∆α, αi = hdα, dαi + hδα, δαi, so the result follows.



8.5.3 Theorem (The Hodge Decomposition Theorem). Let M be a compact, boundaryless, oriented, Riemannian manifold and let ω ∈ Ωk (M ). Then there is an α ∈ Ωk−1 (M ), β ∈ Ωk+1 (M ), and γ ∈ Ωk (M ) such that ω = dα + δβ + γ, where ∆(γ) = 0. Furthermore, dα, δβ, and γ are mutually L2 orthogonal and so are uniquely determined. That is, Ωk (M ) = dΩk−1 (M ) ⊕ δΩk+1 (M ) ⊕ Hk .

(8.5.1)

We can easily check that the spaces in the Hodge decomposition are orthogonal. For example, dΩk−1 (M ) and δΩk+1 (M ) are orthogonal since hdα, δβi = hddα, βi = 0, δ being the adjoint of d and d2 = 0. The basic idea behind the proof of the Hodge theorem can be abstracted as follows. We consider a linear operator T on a Hilbert space E and assume that T 2 = 0. In our case T = d and E is the space of L2 forms. (We ignore the fact that T is only densely defined.) Let T ∗ be the adjoint of T . Let H = { x ∈ E | T x = 0 and T ∗ x = 0 } .

8.5 Introduction to Hodge–deRham Theory

511

We assert that E = cl(range T ) ⊕ cl(range T ∗ ) ⊕ H

(8.5.2)

which, apart from technical points on understanding the closures, is the essential content of the Hodge decomposition. To prove equation (8.5.2), note that the ranges of T and T ∗ are orthogonal because hT x, T ∗ yi = hT x2 , yi = 0. If C denotes the orthogonal complement of cl(range T ) ⊕ cl(range T ∗ ), then H ⊂ C. If x ∈ C then hT y, xi = 0 for all y implies T ∗ x = 0. Similarly, T x = 0, so C ⊂ H and hence C = H. The complete proof of the Hodge theorem requires elliptic estimates and may be found in Morrey [1966]. For more elementary expositions, consult Flanders [1963] and Warner [1983]. 8.5.4 Corollary. Let Hk denote the space of harmonic k-forms. Then the vector spaces Hk and H k (= ker dk / range dk−1 ) are isomorphic. Proof. Map Hk → ker dk by inclusion and then to H k by projection. We need to show that this map is an isomorphism. Suppose γ ∈ Hk and [γ] = 0 where [γ] ∈ H k is the class of γ. But [γ] = 0 means that γ is exact; γ = dβ. But since δγ = 0, γ is orthogonal to dβ; that is, γ is orthogonal to itself, so γ = 0. Thus the map γ 7→ [γ] is one-to-one. Next let [ω] ∈ Hk . We can, by the Hodge theorem, decompose ω = dα + δβ + γ, where γ ∈ Hk . Since dω = 0, dδβ = 0, so 0 = hβ, dδβi = hδβ, δβi, so δβ = 0. Thus, ω = dα + γ and hence [ω] = [γ], so the map γ 7→ [γ] is onto.  The space Hk ∼ = H k is finite dimensional. Again the proof relies on elliptic theory (the kernel of an elliptic operator on a compact manifold is finite dimensional). The Hodge theorem plays a fundamental role in incompressible hydrodynamics, as we shall see in §8.2. It allows the introduction of the pressure for a given fluid state. It has applications to many other areas of mathematical physics and engineering as well; see for example, Fischer and Marsden [1979] and Wyatt, Chua, and Oster [1978]. Below we shall state a generalization of the Hodge theorem for some decomposition theorems for general elliptic operators (rather than the special case of the Laplacian). However, we first pause to discuss what happens if a boundary is present. This theory was worked out by Kodaira [1949], Duff and Spencer [1952], and Morrey [1966, Chapter 7]. Differentiability across the boundary is very delicate, but important. Some of the best results in this regard are due to Morrey. Note that d and δ may not be adjoints in this case, because boundary terms arise when we integrate by parts (see Exercise 8.5-5). Hence we must impose certain boundary conditions. Let α ∈ Ωk (M ). Then α is called parallel or tangent to ∂M if the normal part, defined by nα = i∗ (∗α) is zero where i : ∂M → M is the inclusion map. Analogously, α is perpendicular or normal to ∂M if its tangent part, defined by tα = i∗ (α) is zero. Let X be a vector field on M . Using the metric, we know when X is tangent or perpendicular to ∂M . Now X corresponds to the one–form X [ and also to the (n − 1)-form iX µ = ∗X [ (µ denotes the Riemannian volume form). One checks that X is tangent to ∂M if and only if X [ is tangent to ∂M iff iX µ is normal to ∂M . Similarly X is normal to ∂M iff iX µ is tangent to ∂M . Set Ωkt (M ) = { α ∈ Ωk (M ) | α is tangent to ∂M }, Ωkn (M ) = { α ∈ Ωk (M ) | α is perpendicular to ∂M }, and Hk (M ) = { α ∈ Ωk (M ) | dα = 0, δα = 0 }. The condition that dα = 0 and δα = 0 is, in general, stronger than ∆α = 0 when M has a boundary. One calls elements of Hk harmonic fields, after Kodaira [1949].

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8. Integration on Manifolds

8.5.5 Theorem (Hodge Theorem for Manifolds with Boundary). Let M be a compact oriented Riemannian manifold with boundary. The following decomposition holds: k Ωk (M ) = dΩk−1 (M ) ⊕ δΩk+1 t n (M ) ⊕ H .

One can easily check from the following formula (obtained from Stokes’ theorem): Z hdα, βi = hα, δβi + α ∧ ∗β ∂M

(see Exercise 8.5-5), that the summands in this decomposition are orthogonal. Two other closely related decompositions of interest are (i) Ωk (M ) = dΩk−1 (M ) ⊕ Dtk where Dtk = { α ∈ Ωkt (M ) | δα = 0 } are the co-closed k-forms tangent to ∂M and, dually (ii) Ωk (M ) = δ(Ωk+1 (M )) ⊕ Cnk where Cnk are the closed k-forms normal to ∂M . To put the Hodge theorem in a general context, we give a brief discussion of differential operators and their symbols. (See Palais [1965a], Wells [1980], and Marsden and Hughes [1983] for more information and additional details on proofs.) Let E and F be vector bundles of M and let C ∞ (E) denote the C ∞ sections of E. Assume M is Riemannian and that the fibers of E and F have inner products. A kth order differential operator is a linear map D : C ∞ (E) → C ∞ (F ) such that, if f ∈ C ∞ (E) and f vanishes to kth order at x ∈ M , then D(f )(x) = 0. It is not difficult to see that vanishing to kth order makes intrinsic sense independent of charts and that D is a kth order differential operator iff in local charts D has the form X

D(f ) =

αj

0≤|j|≤k

∂ |j| f , . . . ∂xjs

∂xj1

where j = (j1 , . . . , js ) is a multi-index and αj is a C ∞ matrix-valued function of x (the matrix corresponding to linear maps of E to F ). The operator D has an adjoint operator D∗ given in charts (with the standard Euclidean inner product on fibers) by X 1 ∂ |j| D∗ (h) = (−1)|j| (ραjt h), ρ ∂xj1 . . . ∂xjs 0≤|j|≤k

where ρ dx1 ∧ · · · ∧ dxn is the volume element on M and αjt is the transpose of aj . The crucial property of D∗ is hg, D∗ hi = hD g, hi, where h , i denotes the L2 inner product, g ∈ Cc∞ (E), and h ∈ Cc∞ (F ). That is, g and h are C ∞ sections with compact support. For example, we have the differential operators d : C ∞ (Λk ) → C ∞ (Λk+1 )

(first order)

δ : C ∞ (Λk ) → C ∞ (Λk−1 )

(first order)

∞

k

∞

k

∆ : C (Λ ) → C (Λ )

(second order)

where d∗ = δ, δ ∗ = d, and ∆∗ = ∆. The symbol of D assigns to each ξ ∈ Tx∗ M , a linear map σ(ξ) : Ex → Fx defined by   1 σ(ξ)(e) = D (g − g(x))k f (x), k!

8.5 Introduction to Hodge–deRham Theory

513

where g ∈ C ∞ (M, R), dg(x) = ξ, and f ∈ C ∞ (E), f (x) = e. By writing this out in coordinates one sees that σ(ξ) so defined is independent of g and f and is a homogeneous polynomial expression in ξ of degree k obtained by substituting each ξj in place of ∂/∂xj in the highest order terms. For example, if D(f ) =

X

αij

∂2f + (lower order terms), ∂xi ∂xj

then σ(ξ) =

X

αij ξi ξj

(αij is for each i, j a map of Ex to Fx ). For real-valued functions, the classical definition of an elliptic operator is that the foregoing quadratic form be definite. This can beVgeneralized V as follows: D is called k k elliptic if σ(ξ) is an isomorphism for each ξ = 0. To see that ∆ : C ∞ ( ) → C ∞ ( ) is elliptic one uses the following facts: 1. The symbol of d is σ(ξ) = ξ ∧ . 2. The symbol of δ is σ(ξ) = iξ# . 3. The symbol is multiplicative: σ(ξ)(D1 ◦ D2 ) = σ(ξ)(D1 ) ◦ σ(ξ)(D2 ). From these, it follows by a straightforward calculation that the symbol of ∆ is given by σ(ξ)α = kξk2 α, so ∆ is elliptic. (Compute ξ ∧ (iξ# α) + iξ# (ξ ∧ α) applied to (v1 , . . . , vk ), noting that all but one term cancel.) 8.5.6 Theorem (Elliptic Splitting Theorem—Fredholm Alternative). above. Then C ∞ (F ) = D(C ∞ (E)) ⊕ ker(D∗ )

Let D be an elliptic operator as

Indeed this holds if it is merely assumed that either D or D∗ has injective symbol. The proof of this leans on elliptic estimates that are not discussed here. As in the Hodge theorem, the idea is that the L2 orthogonal complement of range D is ker(D)∗ . This yields an L2 splitting and we get a C ∞ splitting via elliptic estimates. The splitting in case D (resp., D∗ ) has injective symbol relies on the fact that then D∗ D (resp., D D∗ ) is elliptic. For example, the equation D u = f is soluble iff fR is orthogonal to ker(D∗ ). More specifically, ∆u = f is soluble if f is orthogonal to the constants; that is, f dµ = 0. The Hodge theorem is derived from the elliptic splitting theorem as follows. Since ∆ is elliptic and symmetric Vk C ∞ ( (M )) = range(∆) ⊕ ker(∆) = range(∆) ⊕ H Now write a k-form ω as ω = ∆ρ + γ = dδρ + δdρ + γ, so to get Theorem 8.5.3, we can choose α = δρ and β = dρ.

Supplement 8.5A Introduction to Degree Theory One of the purposes of degree theory is to provide algebraic measures of the number of solutions of nonlinear equations. Its development rests on Stokes’ theorem. It beautifully links calculus on manifolds with ideas on differential and algebraic topology. All manifolds in this section are assumed to be finite dimensional, paracompact and Lindel¨ of. We begin with an extendability result.

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8. Integration on Manifolds

8.5.7 Proposition. Let V and N be orientable manifolds, dim(V ) = n+1 and dim(N ) = n. If f : ∂V → N is a smooth proper map that extends to a smooth map of V to N , then for every ω ∈ Ωn (N ) with compact support, Z f ∗ ω = 0.

∂V

Proof.

Let F : V → N be a smooth extension of f. Then by Stokes’ theorem Z Z Z Z ∗ ∗ ∗ f ω= F ω= dF ω = F ∗ dω = 0. ∂V

∂V

V

V

since dω = 0.



This proposition will be applied to the case V = [0, 1] × M . For this purpose let us recall the product orientation (see Exercise 7.5-14). If N and M are orientable manifolds (at most one of which has a boundary), then N ×M is a manifold (with boundary), which is orientable in the following way. Let π1 : N ×M → N and π2 : N × M → M be the canonical projections and [ω], [η] orientations on N and M respectively. Then the orientation of N × M is defined to be [π1∗ ω ∧ π1∗ η]. Alternatively, if v1 , . . . , vn ∈ Tx N and w1 , . . . , wm ∈ Ty M are positively oriented bases in the respective tangent spaces, then (v1 , 0), . . . , (vn , 0), (0, w1 ), . . . , (0, wm ) ∈ T(x,y) (N × M ) is defined to be a positively oriented basis in their product. Thus, for [0, 1] × M, a natural orientation will be given at every point (t, x) ∈ [0, 1] × M by (1, 0), (0, v1 ), . . . , (0, vm ), where v1 , . . . , vm ∈ Tx M is a positively oriented basis. The boundary orientation of [0, 1] × M is determined according to Definition 8.2.7 Since ∂([0, 1] × M ) = ({0} × M ) ∪ ({1} × M ), every element of this union is oriented by the orientation of M . On the other hand, this union is oriented by the boundary orientation of [0, 1] × M . Since the outward normal at (1, x) is (1, 0), we see that a positively oriented basis of T(1,x) ({1} × M ) is given by (0, v1 ), . . . , (0, vm )

for v1 , . . . , vm ∈ Tx M,

a positively oriented basis. However, since the outward normal at (0, x) is (−1, 0), a positively oriented basis of T(0,x) ({0} × M ) must consist of elements (0, w1 ), . . . , (0, wm ) such that (−1, 0), (0, w1 ), . . . , (0, wm ) is positively oriented in [0, 1] × M, that is, defines the same orientation as (1, 0), (0, v1 ), . . . , (0, vm ),

for v1 , . . . , vm ∈ Tx M

a positively oriented basis. This means that w1 , . . . , wm ∈ Tx M is negatively oriented (see Figure 8.5.1). Thus the oriented manifold ∂([0, 1] × M ) is the disjoint union of {0} × M , where M is negatively oriented, with f {1} × M , where M is positively oriented. 8.5.8 Definition. Two smooth mappings f, g : M → N , are called C r -homotopic if there is a C r -map F : [0, 1] × M → N such that F (0, m) = f (m)

and

F (1, m) = g(m),

for all m ∈ M . The homotopy F is called proper if F is a proper map; in this case f and g are said to be properly C r -homotopic maps. Note that if f and g are properly homotopic then necessarily f and g are proper as restrictions of a proper map to the closed sets {0} × M and {1} × M , respectively.

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515

Figure 8.5.1. Orientation on spheres

8.5.9 Proposition. Let M and N be orientable n-manifolds, with M boundaryless, let ω ∈ Ωn (N ) have compact support, and suppose f, g : M → N are properly smooth homotopic maps. Then Z

f ∗ω =

M

Proof.

Z

g ∗ ω.

M

There are two ways to do this.

Method 1. Let F : [0, 1] × M → N be the proper homotopy between f and g. By the remarks preceding Definition 8.5.8, we have Z Z Z Z f ∗ω = − f ∗ ω and g∗ ω = g ∗ ω, {0}×M

so that

Z M

g∗ ω −

{1}×M

M

Z

f ∗ω =

M

Z

M

(F |∂([0, 1] × M ))∗ ω = 0

∂([0,1]×M )

by Proposition 8.5.7. Method 2. By Theorem 7.4.16, f ∗ ω − g ∗ ω = dη for some η ∈ Ωn−1 (M ) which has compact support since the homotopy between f and g is a proper map. Then by Stokes’ theorem Z Z Z f ∗ω − g∗ ω = dη = 0.  M

M

M

Remark. Properness of f and g does not suffice in the hypothesis of Proposition 8.5.9. For example, if M = N = R, ω = adx with α ≥ 0 a C ∞ function satisfying supp(a) ⊂ ]−∞, 0[, then f (x) = x and g(x) = x2 R +∞ R +∞ are smoothly but not properly homotopic via F (t, x) = (1 − t)x + tx2 and −∞ f ∗ ω > 0, while −∞ g ∗ ω = 0 since g ∗ ω = 0.  8.5.10 Theorem (Degree Theorem). Let M and N be oriented n-manifolds, N connected, M boundaryless, and f : M → N a smooth proper map. Then there is an integer deg(f ) constant on the proper homotopy class of f , called the degree of f such that for any η ∈ Ωn (N ) with compact support, Z Z f ∗ η = deg(f )

M

η. N

(8.5.3)

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8. Integration on Manifolds

If x ∈ M is a regular point of f , let sign(Tx f ) be 1 or −1 depending on whether the isomorphism Tx f : Tx M → Tf (x) N preserves or reverses orientation. The integer deg(f ) is given by X deg(f ) = sign(Tx f ). (8.5.4) x∈f −1 (y)

where y is an arbitrary regular value of f ; if y ∈ / f (M ) the right hand side is by convention equal to zero. R Proof. By Proposition 8.5.9, M f ∗ η depends only on the proper homotopy class of f (and on η). By Sard’s theorem, there is a regular value y of f . There are two possibilities: either Rf = N \f (M ) or not. If Rf = N \f (M ), then Tx f is never onto for all x ∈ M . For any v1 , . . . , vn ∈ Tx M, (f ∗ η)(x)(v1 , . . . , vn ) = η(f (x))(Tx f (v1 ), . . . , Tx f (vn )) = 0, since Tx f (v1 ), . . . , Tx f (vn ) are linearly dependent. Thus deg(f ) exists and equals zero. Assume Rf ∩ f (M ) 6= ∅ and let y ∈ Rf ∩ f (M ). Since M and N have the same dimension, f −1 (y) is a zero-dimensional submanifold of M , hence discrete. Properness of f implies that f −1 (y) is also compact, that is, f −1 (y) = {x1 , . . . , xk+l }, where Txi f is orientation preserving for i = 1, . . . , k and orientation reversing for i = k + 1, . . . , k + l. The inverse function theorem implies that there are open neighborhoods Ui of xi and V of y such that f −1 (V ) = U1 ∪ · · · ∪ Uk+l , Ui ∩ Uj = ∅ and if f |Ui : Ui → V is a diffeomorphism. If supp(η) ⊂ V , then by the change of variables formula Z

∗

f η= M

k+l Z X

Z

∗

f η = (k − l)

Ui

i=1

Z η = (k − l)

V

η

(8.5.5)

N

and so the theorem is proved for supp(η) ⊂ V . To deal with a general η proceed in the following way. For the open neighborhood V of η, consider the collection of open subsets of N , S = { ϕ(V ) | ϕ is a diffeomorphism properly homotopic to the identity }. We shall prove that S covers N . Let n ∈ N ; we will show that there is a diffeomorphism ϕ properly homotopic to the identity such that ϕ(n) = y. Let c : [0, 1] → N be a smooth curve with c(0) = n and c(1) = y. As in Theorem 6.5.9, use a partition of unity to extend c0 (t) to a smooth vector field X ∈ X(N ) such that X vanishes outside a compact neighborhood of c([0, 1]). The flow Ft of X is complete by Corollary 4.1.20 and is the identity outside the above compact neighborhood of c([0, 1]). Thus the restriction F : [0, 1] × N → N is proper. Then ϕ = F1 is a proper diffeomorphism properly homotopic to the identity on N and ϕ(n) = F1 (n) = c(1) = y. Since S covers N , choose a partition of unity {(Vα , hα )} subordinate to S and let ηα = hα η; thus, supp(η) ⊂ Vα ⊂ ϕα (V ) for some ϕα . Since all ϕα are orientation preserving, the change of variables formula and equation (8.5.5) give Z XZ XZ XZ (k − l) η = (k − l) ηα = (k − l) ϕ∗α ηα = f ∗ ϕ∗α ηα . N

Vα

α

α

V

α

M

Since ϕα is properly homotopic to the identity and f is proper, it follows that ϕα ◦ f is properly homotopic to f . Thus by Proposition 8.5.9, Z Z (ϕα ◦ f )∗ ηα = f ∗ ηα M

M

and therefore, Z (k − l)

η= N

XZ α

M

∗

Z

f ηα = M

f ∗ η.



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517

Notice that by construction, if deg(f ) 6= 0, then f is onto, so f (x) = y is solvable for x given y. 8.5.11 Corollary. Let V and N be orientable manifolds with dim(V ) = n + 1, and dim(N ) = n. If f : ∂V → N extends to V , then deg(f ) = 0. This is a reformulation of Proposition 8.5.7. Similarly, Proposition 8.5.9 is equivalent to the following. 8.5.12 Corollary. Let M , N be orientable n-manifolds, N connected, M boundaryless, and let f, g : M → N be smooth properly homotopic maps. Then deg (f ) = deg(g). This corollary is useful in three important applications. The first concerns vector fields on spheres. 8.5.13 Theorem (Hairy Ball Theorem). point.

Every vector field on an even dimensional sphere has a critical

Proof. Let S 2n be embedded as the unit sphere in R2n+1 and X ∈ X(S 2n ). Then X defines a map f : S 2n → R2n+1 with components f (x) = (f 1 (x), . . . , f 2n+1 (x)) satisfying f 1 (x)x1 + · · · + f 2n+1 (x)x2n+1 = 0. Here f i (x) are the components of X in R2n+1 . Assume that X has no critical point. Replacing f by f /kf k, we can assume that f : S 2n → S 2n . The map F : [0, 1] × S 2n → S 2n ,

F (t, x) = (cos πt)x + (sin πt)f (x)

is a smooth proper homotopy between F (0, x) = x and F (1, x) = −x. That is, the identity Id is homotopic to the antipodal map A : S 2n → S 2n , A(x) = −x. Thus by Corollary 8.5.12, deg A = 1. However, since the Jacobian of A is −1 (this is the place where we use evenness of the dimension of the sphere), A is orientation reversing and thus by the Degree theorem 8.5.10, deg(A) = −1, which is a contradiction.  The second application is to prove the existence of fixed points for maps of the unit ball to itself. 8.5.14 Theorem (Brouwer’s Fixed-Point Theorem). itself has a fixed point.

A smooth mapping of the closed unit ball of Rn into

Proof. Let B denote the closed unit ball in Rn and let S n−1 = ∂B be its boundary, the unit sphere. If f : B → B has no fixed point, define g(x) ∈ S n−1 to be the intersection of the line starting at f (x) and going through x with S n−1 . The map g : B → S n−1 so defined is smooth and for x ∈ S n−1 , g(x) = x. If n = 1 this already gives a contradiction, since g must map B = [−1, 1] onto {−1, 1} = S 0 , which is disconnected. For n ≥ 2, define a smooth proper homotopy F : [0, 1] × S n−1 → S n−1 by F (t, x) = g(tx). Thus F is a homotopy between the constant map c : S n−1 → S n−1 , c(x) = g(0) and the identity of S n−1 . But c∗ ω = 0 for any ω ∈ Ωn−1 (S n−1 ), so that by Theorem 8.5.10, deg c = 0. On the other hand, by Corollary 8.5.12, deg(c) = 1, which is false.  The Brouwer fixed point theorem is valid for continuous mappings and is proved in the following way. If f has no fixed points, then by compactness there exists a positive constant K > 0 such that kf (x) − xk > K for all x ∈ B. Let  < min(K, 2) and choose δ > 0 such that 2δ/(1 + δ) < ; that is, δ < /(2 − ). By the Weierstrass approximation theorem (see, for example, Marsden and Hoffman [1993]) there exists a polynomial mapping q : Rn → Rn such that kf (x) − q(x)k < δ for all x ∈ B. The image q(B) lies inside the closed ball centered at 0 of radius 1 + δ, so that p ≡ q/(1 + δ) : B → B and q(x) 2δ f (x) f (x) kf (x) − p(x)k ≤ f (x) − + − ≤ < 1+δ 1+δ 1+δ 1+δ for all x ∈ B. Since p is smooth by Theorem 8.5.14, it has a fixed point, say x0 ∈ B. Then 0 < K < kf (x0 ) − x0 k ≤ kf (x0 ) − p(x0 )k + kp(x0 ) − x0 k ≤ ,

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8. Integration on Manifolds

which contradicts the choice  < K. Brouwer’s fixed point theorem is false in an open ball, for the open ball is diffeomorphic to Rn and translation provides a counterexample. The proof we have given is not “constructive.” For example, it is not clear how to base a numerical search on this proof, nor is it obvious that the fixed point we have found varies continuously with f . For these aspects, see Chow, Mallet-Paret, and Yorke [1978]. A third application of Corollary 8.5.12 is a topological proof of the fundamental theorem of algebra. 8.5.15 Theorem (The Fundamental Theorem of Algebra). has a root.

Any polynomial p : C → C of degree n > 0

Proof. Assume without loss of generality that p(z) = z n + an−1 z n−1 + · · · + a0 , where ai ∈ C, and regard p as a smooth map from R2 to R2 . If p has no root, then we can define the smooth map f (z) = p(z)/|p(z)| whose restriction to S 1 we denote by g : S 1 → S 1 . Let R > 0 and define for t ∈ [0, 1] and z ∈ S 1 ,   pt (z) = (Rz)n + t an−1 (Rz)n−1 + · · · + a0 . Since

  a0 (z) −1 = 1 + t an−1 (Rz) + · · · + pt (Rz)n (Rz)−n

and the coefficient of t converges to zero as R → ∞, we conclude that for sufficiently large R, none of the pt has zeros on S 1 . Thus, F : [0, 1] × S 1 → S 1

defined by F (t, z) =

pt (z) |pt (z)|

is a smooth proper homotopy of dn (z) = z n with g(Rz), which in turn is properly homotopic to g(z). On the other hand, G : [0, 1] × S 1 → S 1 defined by G(t, z) = f (tz) is a proper homotopy of the constant mapping c : S 1 → S 1 , c(z) = f (0) with g. Thus dn is properly homotopic to a constant map and hence deg dn = 0 by Corollary 8.5.12. However, if S 1 is parameterized by arc length θ, 0 ≤ θ ≤ 2π, then dn maps the segment 0 ≤ θ ≤ 2π/n onto the segment 0 ≤ θ ≤ 2π since dn has the effect eiθ 7→ einθ . If ω denotes the corresponding volume form on S 1 , the change of variables formula thus gives Z Z d∗n ω = n ω = 2πn, that is, deg dn = n, S1

S1

which for n 6= 0 is a contradiction.



The fundamental theorem of algebra shows that any polynomial p : C → C of degree n can be written as p(z) = c(z − z1 )k1 · · · (z − zm )km , where z1 , . . . , zm are the distinct roots of p, k1 , . . . , km are their multiplicities, k1 + · · · + km = n, and c ∈ C is the coefficient of z n in p(z). The fundamental theorem of algebra can be refined to take into account multiplicities of roots in the following way. 8.5.16 Proposition. Let D be a compact subset of C with open interior and smooth boundary ∂D. Assume that the polynomial p : C → C has no zeros on ∂D. Then the total number of zeros of p which lie in the interior of D, counting multiplicities, equals the degree of the map p/|p| : ∂D → S 1 . Proof. Let z1 , . . . , zm be the roots of p in the interior of D with multiplicities k(1), . . . , k(m). Around each zi construct an open disk Di centered at zi , Di ⊂ D, such that ∂D ∩ ∂Di = ∅ and ∂Di ∩ ∂Dj = ∅,

8.5 Introduction to Hodge–deRham Theory

519

for all i 6= j. Then V = D\(D1 ∪ · · · ∪ Dm ) is a smooth compact two-dimensional manifold whose boundary is ∂D ∪ ∂D1 ∪ . . . ∂Dm . The boundary orientation of ∂Di induced by V is opposite to the usual boundary orientation of ∂Di as the boundary of the disk Di ; see Figure 8.5.2. Since p/|p| is defined on all of V , Corollary 8.5.11 implies that the degree of p/|p| : ∂V → S 1 is zero. But the degree of a map defined on a disjoint union of manifolds is the sum of the individual degrees and thus the degree of p/|p| on ∂D equals the sum of the degrees of p/|p| on all ∂Dj . The proposition is therefore proved if we show that the degree of p/|p| on ∂Di is the multiplicity k(i) of the root zi .

Figure 8.5.2. Relating degree with numbers of zeros

Let r(z) = c

m Y

(z − zj )k(j) ,

so

p(z) = (z − zi )k(i) r(z)

j=1,j6=i

and the only zero of p(z) in the disk Di is zi . Then ϕ : z ∈ S 1 → zi +Ri z ∈ ∂Di , where Ri is the radius of Di , is a diffeomorphism and therefore the degree of p/|p| : ∂Di → S 1 equals the degree of (p◦ϕ)/|p◦ϕ| : S 1 → S 1 . The homotopy H : [0, 1] × S 1 → S 1 of z k(i) arg(r(zi )) with (p ◦ ϕ)/|p ◦ ϕ| given by H(t, z) =

z k(i) r(zi + tRi z) |r(zi + tRi z)|

is proper and smooth, since zi + tRi z ∈ Di for all z ∈ S 1 , t ∈ [0, 1]. Thus in ∂Di we have deg

p p◦ϕ = deg = deg z k(i) = k(i). |p| |p ◦ ϕ|



A variant of the fundamental theorem of algebra is the following. 8.5.17 Proposition. Let U be an open subset of Rn and f : U → Rn be a C 1 proper map. Assume there is a closed subset K ⊂ U such that for all x ∈ U \K, the Jacobian J(f )(x) does not change sign and is not identically equal to zero. Then f is surjective. Proof. The map f cannot be constant since the Jacobian J(f )(x) is not identically zero for all x ∈ U \K. For the same reason, f has a regular value y ∈ f (U \K), for if all values in f (U \K) are singular, J(f ) will vanish on U \K. If y ∈ f (U \K) is a regular value of f then sign(Tx f ) does not change for all x ∈ f −1 (y) so by the degree theorem 8.5.10, deg(f ) 6= 0, which then implies that f is onto. 

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8. Integration on Manifolds

The orientation preserving character of proper diffeomorphisms is characterized in terms of the degree as follows. 8.5.18 Proposition. Let M and N be oriented boundaryless connected manifolds and f : M → N a proper local diffeomorphism. Then deg f = 1, if and only if f is an orientation preserving diffeomorphism. Proof. If f is an orientation preserving diffeomorphism, then deg(f ) = 1 by Theorem 8.5.10. Conversely, let f be a proper local diffeomorphism with deg f = 1. Define U± = { m ∈ M | sign Tm f = ±1 } . Since f is a local diffeomorphism, U± are open in M . Connectedness of M and M = U+ ∪ U− ,

U+ ∩ U− = ∅

imply that M = U+ or M = U− . Let us show that U− = ∅. Since deg(f ) = 1, f is onto and hence if n ∈ N , f −1 (n) 6= ∅ is a discrete submanifold of M . Properness of f implies that f −1 (n) = {m(1), . . . , m(k)}. Since f is a local diffeomorphism of a neighborhood Ui of m(i) onto a neighborhood V of n, sign Tm(i) f is the same for all i = 1, . . . , k (for otherwise J(f ) must vanish somewhere). Thus deg(f ) = ±k according to whether Tm(i) f preserves or reverses orientation. Since deg(f ) = 1, this implies U− = ∅ and k = 1, that is, f is injective. Thus f is a bijective local diffeomorphism, that is, a diffeomorphism. 

Supplement 8.5B Zero and n-Dimensional Cohomology Here we compute H 0 (M ) and H n (M ) for a connected n-manifold M . Recall that the cohomology groups are defined by H k (M ) = ker(d)k / range(dk−1 ), where dk : Ωk (M ) → Ωk+1 (M ) is the exterior differential. If Ωkc (M ) denotes the k-forms with compact sup(M ) and one forms in the same manner Hck (M ), the compactly supported port, then dk : Ωkc (M ) → Ωk+1 c cohomology groups of M . Thus, H 0 (M ) = { f ∈ F(M ) | df = 0 } ∼ =R since any locally constant function on a connected space is constant. If M were not connected, then H 0 (M ) = Rc , where c is the number of connected components of M . By the Poincar´e lemma, if M is contractible, then H q (M ) = 0 for q 6= 0. The rest of this supplement is devoted to the proof and applications of the following special case of deRham’s theorem. 8.5.19 Theorem.

Let M be a boundaryless connected n-manifold.

(i) If M is orientable, then Hcn (M ) R∼ = R, the isomorphism being given by integration: [ω] 7→ particular ω ∈ Ωnc (M ) is exact iff M ω = 0. (ii) If M is nonorientable, then Hcn (M ) = 0.

R M

ω. In

8.5 Introduction to Hodge–deRham Theory

521

(iii) If M is non-compact, orientable or not, then H n (M ) = 0. R Before starting the actual proof, let us discuss (i). The integration mapping M : Ωn (M ) → R is linear and onto. To see that it is onto, let ω be an n-form Rwith support in a chart in which the local expression is R ω = f dx1 ∧ · · · ∧ dxn with f a bump function. Then M ω = Rn f (x)dx > 0. Since we can multiply ω by any scalar, the integration map is onto. Any ω with nonzero integral cannot be exact by Stokes’Rtheorem. This last remark also shows that integration induces a mapping, which we shall still call integration, M : Hcn (M ) → R, which is linear and onto. Thus, in order to show that itR is an isomorphism as (i) states, it is necessary and sufficient to prove it is injective, that is, to show that if M ω = 0 for ω ∈ Ωn (M ), then ω is exact. The proof of this will be done in the following lemmas. 8.5.20 Lemma.

Theorem 8.5.19 holds for M = S 1 .

Proof. Let p : R → S 1 be given by p(t) = eit and ω ∈ Ω1 (S 1 ). Then p∗ ω = f dt for f ∈ F(R) a 2π-periodic function. Let F be an antiderivative of f . Since Z 0=

Z

S1

t+2π

f (s)ds = F (t + 2π) − F (t)

ω= t

for all t ∈ R, we conclude that F is also 2π-periodic, so it induces a unique map G ∈ F(S 1 ), determined by p∗ G = F . Hence p∗ ω = dF = p∗ dG implies ω = dG since p is a surjective submersion.  8.5.21 Lemma.

Theorem 8.5.19 holds for M = S n , n > 1.

Proof. This will be done by induction on n, the case n = 1 being the previous lemma. Write S n = N ∪ S, where N = { x ∈ S n | xn+1 ≥ 0 } is the closed northern hemisphere and S = { x ∈ S n | xn+1 ≤ 0 } the closed southern hemisphere. Then N ∩ S = S n−1 is oriented in two different ways as the boundary of N and S, respectively. Let ON =



x ∈ S n | xn+1 > − ,

OS =



x ∈ S n | xn+1 < 



be open contractible neighborhoods of N and S, respectively. Thus by the Poincar´e lemma, there exist αN ∈ Ωn−1 (ON ), αS ∈ Ωn−1 (OS ) such that dαN = ω on ON , dαS = ω on OS . Hence by hypothesis and Stokes’ theorem, Z Z Z Z Z Z Z ∗ 0= ω= ω+ ω= dαN + dαS = i αN + i∗ αS Sn N S N S ∂N ∂S Z Z = i∗ αN − i∗ αS S n−1 S n−1 Z = i∗ (αN − αS ) S n−1

where i : S n−1 → S n is the inclusion of S n−1 as the equator of S n ; the minus sign appears on the second integral because the orientations of S n−1 and ∂S are opposite. By induction, i∗ (αN − αS ) ∈ Ωn−1 (S n−1 ) is exact. Let O = ON ∩OS and note that the map r : O → S n−1 , sending each x ∈ S to r(x) ∈ S n−1 , the intersection of the meridian through x with the equator S n−1 , is smooth. Then r ◦ i is the identity on S n−1 . Also, i ◦ r is homotopic to the identity of O, the homotopy being given by sliding x ∈ O along the meridian to r(x). Since d(αN − αS ) = ω − ω = 0 on O, by Theorem 7.4.16 we conclude that (αN − αS ) − r∗ i∗ (αN − αS ) is exact on O. But we just showed that i∗ (αN − αS ) ∈ Ωn−1 (S n−1 ) is exact, and hence r∗ i∗ (αN − αS ) ∈ Ωn−1 (O) is also exact. Hence αN − αS ∈ Ωn−1 (O) is exact. Thus, there exists β ∈ Ωn−2 (O) such that αN − αS = dβ

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8. Integration on Manifolds

on O. Now use a bump function to extend β to a form γ ∈ Ωn−2 (S n ) so that on O, β = γ, and γ = 0 on S n \V , where V is an open set such that cl(U ) ⊂ V . Then ( αN (x), if x ∈ N, λ(x) = αS (x) + dγ, if x ∈ S is by construction C ∞ and dλ = ω.



8.5.22 Lemma. A compactly supported n-form ω ∈ Ωn (Rn ) is the exterior derivative of a compactly R n supported (n − 1)-form on R iff Rn ω = 0. Proof. Let σ : S n → Rn be the stereographic projection from the north pole (0, . . . , 1) ∈ S n onto Rn and assume without loss of generality that (0, . . . , 1) ∈ / σ −1 (supp ω). By the previous lemma, σ ∗ ω = dα, for n−1 n some α ∈ Ω (S ) since Z Z 0=

σ∗ ω

ω= Rn

S

by the change of variables formula. But σ ∗ ω = dα is zero in a contractible neighborhood U of the north pole, so that by the Poincar´e lemma, α = dβ on U , where β ∈ Ωn−2 (U ). Now extend β to an (n − 2)-form γ ∈ Ωn−2 (S n ) such that β = γ on U and γ = 0 outside a neighborhood of cl(U ). But then σ∗ (α − dγ) is compactly supported in Rn and dσ∗ (α − dγ) = σ∗ dα = ω.  8.5.23 Lemma.

Let M be a boundaryless connected n-manifold. Then Hcn (M ) is at most one-dimensional.

n Proof. Let (U0 , ϕ0 ) be a chart on M such that ϕ0 (U0 ) is the open unit ball B in Rn . Let R ω ∈ Ωc (M ), 1 n n satisfying supp(ω) ⊂ U1 , be the pull–back of a form f dx ∧· · ·∧dx ∈ Ω (B) where f ≥ 0 and Rn f (x)dx = 1. To prove the lemma, it is sufficient to show that for every η ∈ Ωnc (M ) there exists a number c ∈ R such that (M ). η − cω = dζ for some ζ ∈ Ωn−1 c (M ) has supp(η) entirely contained in a chart (U, ϕ) and let U0 , U1 , . . . , Uk be a First assume η ∈ Ωn−1 c finite covering of a curve starting in U0 and ending on Uk = U such that Ui ∩ Ui+1 = ∅. Let αi ∈ Ωnc (Ui ), i = 1, . . . , k − 1 be non-negative n-forms such that Z supp(αi ) ⊂ Ui , supp(αi ) ∩ Ui+1 6= ∅, and ϕi∗ (αi ) = 1. Rn

Let α0 = ω and αk = η. But then Z ϕi∗ (αi−1 ) 6= 0,

i = 1, . . . , k

Rn

R R by the change-of-variables formula, so that with ci = −1/ Rn ϕi∗ (αi−1 ) we have Rn ϕi∗ (αi ci αi−1 ) = 0. Thus by the previous lemma ϕi∗ (αi − ci αi−1 ) is the differential of an (n − 1)-form supported in B. That is, there exists βi ∈ Ωn−1 (M ), βi vanishing outside Ui such that c αi − ci αi−1 = dβi ,

i = 1, . . . , k.

Put c = ck · · · c1 and β = βk + (ck βk−1 ) + (ck ck−1 βk−2 ) + · · · + (ck−1 · · · c2 β1 ) ∈ Ωn−1 (M ). Then η − cω = αk − cα0 = αk − ck αk−1 + ck (αk−1 − ck−1 αk−2 ) + . . . + (ck · · · c2 )(α1 − c1 α0 ) = dβk + ck dβk−1 + · · · + (ck · · · c2 )dβ1 = dβ.

8.5 Introduction to Hodge–deRham Theory

523

Let η ∈ Ωnc (M ) be arbitrary and { χi | i = 1, . . . , k } a partition of unity subordinate to the given atlas { (Ui , ϕi ) | i = 1, . . . , k }. Then χi η is compactly supported in Ui and hence there exist constants ci and forms αi ∈ Ωn−1 (M ) such that χi η − ci ω = dαi . If c c=

k X

ci

and α =

i=1

k X

αi ∈ Ωn−1 (M ), c

i=1

then η − cω =

k X

(χi η − ci ω) =

i=1

k X

dαi = dα.



i=1

Proof of Theorem 8.5.19. (i) By the preceding lemma, Hcn (M ) is zero- or one-dimensional. We have R n seen that M : Hc (M ) → R is linear and onto so that necessarily Hcn (M ) is one-dimensional; that is, R ω = 0 iff ω is exact. M f be the oriented double covering of M and π : M f → M the canonical projection. Define π # : (ii) Let M f) by π # [α] = [π ∗ α]. We shall first prove that π # is the zero map. Let {Ui } be an open H n (M ) → H n (M covering of M by chart domains and {χi } a subordinate partition of unity. Let π −1 (U i) = Ui1 ∪ Ui2 . Then f by chart domains and the maps ψ j = χi ◦ π/2 : U j → R, j = 1, 2, { Uij | j = 1, 2 } is an open covering of M i i form a subordinate partition of unity on M . Let α ∈ Ωnc (M ). Then ! Z Z Z k X XZ 1 ∗ 2 ∗ ∗ 1 ∗ ψi π α + ψi π α = 0, π α= ψi π α = f M

i,j

Uij

i=1

Ui1

Ui2

each term vanishing since their push–forwards by the coordinate maps coincide on Rn and Ui1 and Ui2 have f); that is, π # [α] = [π ∗ α] = [0] opposite orientations. By (i), we conclude that π ∗ α = dβ for some β ∈ Ωn−1 (M n for all [α] ∈ Hc (M ). We shall now prove that π # is injective, which will show that Hcn (M ) = 0. Let α ∈ Ωnc (M ) be such that f) and let r : M f→M f be the diffeomorphism associating to (m, [ω]) ∈ M f π ∗ α = dβ for some β ∈ Ωn−1 c(M f the point (m, [−ω]) ∈ M . Then clearly π ◦ r = π so that d(r∗ β) = r∗ (dβ) = r∗ π ∗ α = (π ◦ r)∗ α = π ∗ α = dβ. Define γ˜ ∈ Ωn−1 (M ) by setting γ˜ = (1/2)(β + r∗ β) and note that r∗ γ˜ = γ˜ and c d˜ γ=

dβ + dr∗ β = dβ = π ∗ α. 2

But γ˜ projects to a well-defined form γ ∈ Ωn−1 (M ) such that π ∗ γ = γ˜ , since r∗ γ˜ = γ˜ . Thus π ∗ α = d˜ γ= c ∗ ∗ dπ γ = π dγ, which implies that α = dγ, since π is a surjective submersion. (iii) Assume first that ω ∈ Ωnc (M ) has its support contained in a relatively compact chart domain U1 of M . Then out of a finite open relatively compact covering of cl(U1 ) by chart domains, pick a relatively compact chart domain U2 which does not intersect supp(ω). Working with cl(U2 )\U1 , find a relatively compact chart domain U3 such that U1 ∩ U3 = ∅,

U2 ∩ U3 6= ∅,

U3 ∩ (M \(U1 ∩ U2 )) 6= ∅.

Proceed inductively to find a sequence {Un } of relatively compact chart domains such that Un ∩ Un+1 6= ∅,

Un ∩ Un−1 6= ∅,

Un ∩ Um = ∅

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8. Integration on Manifolds

Figure 8.5.3. A chain of open sets

for all m 6= n − 1, n, n + 1, and such that supp(ω) ⊂ U1 , U2 ∩ supp(ω) = ∅. Since M is not compact, this sequence can be chosen to be infinite; see Figure 8.5.3. Now choose in each Un ∩ Un+1 an n-form ωn with compact support such that Z Z Z Z ω= ω1 = ω2 = · · · = ωn = · · · . U1

U1

U2

Un

Since Hcn (Un ) = R by (i), Un being orientable, ωn−1 and ωn define the same cohomology class, that is, there is ηn ∈ Ωn−1 c(Un ) such that ωn−1 = ωn + dηn . If we let ω0 = ω, we get recursively ! n X ω = dη1 + ω1 = d(η1 + η2 ) + ω2 = · · · = d ηi + ωn = · · · . i=1

P We claim that ω = d( n≥1 S ηn ), where the sum is finite since any point of the manifold belongs to at most two Un ’s. Thus, if p ∈ n≥1 Un , let p ∈ Un so that d(Σn≥1 ηn )(p) = dηn−1 (p) + dηn (p) + dηn+1 (p) = dηn−1 (p) + ωn−1 (p) − ωn+1 (p) = dηn−1 (p) + ωn−1 (p) − ωn+2 (p) − dηn+2 (p) with the convention η0 = 0. Since Un ∩ Un+2 = ∅ and supp ωn+2 , supp ηn+2 ⊂ Un+2 , it follows that the last two terms vanish. Thus,   X d ηn  (p) = dηn−1 (p) + ωn−1 (p). n≥1

If n = 1, this proves the desired equality. If n ≥ 2, then   X d ηn  (p) = dηn−1 (p) + ωn−1 (p) = ωn−2 (p) n≥1

P and Un ∩ Un−2 = ∅ implies S that d( n≥1 ηn ) = 0. Since also ω(p) = 0 in this case, the desired equality holds again. / n≥1 both sides of the equality are zero and we showed that ω is exact, P Finally, if p ∈ P Un , then S ω = d( n≥1 ηn ), with supp( n≥1 ηn ) ⊂ n≥1 Un .

8.5 Introduction to Hodge–deRham Theory

525

Now if ω ∈ Ωn (M ), let {(Ui , gi )} be a partition of unity subordinate to a locally finite atlas of M whose chart domains are relatively compact. Thus supp(gi ω) ⊂ Ui and by what we just proved, gi ω = dηi , with supp(ηi ) contained in the union of the chain of open sets {Uni }, U1i = Ui , as described above. Refine each such chain, such that all its elements are one of the Uj ’s. Since at most P two of the Uni intersect for n−1 each P fixed i, itPfollows that the sum Σi ηi is locally finite and therefore h = (M ). Finally, i ηi ∈ Ω n ω = i gi ω = i dηi = dη, thus showing that ω is exact and hence H (M ) = 0.  One can use this result as an alternative method to introduce the degree of a proper map f : M → N between oriented n-manifolds; that is, that integer deg(f ) such that Z Z ∗ f η = deg(f ) η M n for any R η ∈∗ Ωc (Nn). Indeed, [η] 7→ M f η of Hc (N ) to R

N

∼ since the isomorphism = R is given by [η] 7→ must be some real multiple of this isomorphism: Z Z ∗ f η = deg(f ) η Hcn (N )

M

R N

η, the linear map

N

Ωnc (N )

and some real deg(f ). for all η ∈ To prove that deg(f ) is an integer in this context and that the formula (8.5.3) for deg(f ) is independent of the regular value y, note that if y is any regular value of f and x ∈ f −1 (y), then there exist compact neighborhoods V of y and U of x such that f |U : U → V is a diffeomorphism. Since f −1 (y) is compact and discrete, it must be finite, say f −1 (y) = {x1 , . . . , xk }. This shows that f −1 (V ) = U1 ∪ · · · ∪ Uk with all Ui disjoint and the sum in the degree formula is finite. Shrink V if necessary to lie in a chart domain. Now choose η ∈ Ωnc (N ) satisfying supp(η) ⊂ V . Then   Z  X Z X Z ∗ ∗ f η= f η= sign(Txi f ) η   N M Ui −1 −1 xi ∈f

(y)

xi ∈f

(y)

by the change of variables formula in Rn , so the claim follows. Degree theory can be extended to infinite dimensions as well and has important applications to partial differential equations and bifurcations. This theory is similar in spirit to the above and was developed by Leray and Schauder in the 1930s. See Chow and Hale [1982], Nirenberg [1974], and Elworthy and Tromba [1970b] for modern accounts.

Exercises  8.5-1 (Poincar´e duality).

Show that ∗ induces an isomorphism ∗ : H k → H n−k and Hck → Hcn−k .

 8.5-2. (For students knowing some algebraic topology.) Develop some basic properties of deRham cohomology groups such as homotopy invariance, exact sequences, Mayer–Vietoris sequences and excision. Use this to compute the cohomology of some standard simple spaces (tori, spheres, projective spaces).  8.5-3. (i) Show that any smooth vector field X on a compact Riemannian manifold (M, g) can be written uniquely as X = Y + grad p where Y has zero divergence (and is parallel to ∂M if M has boundary). (ii) Show directly that the equation ∆p = − div X,

(grad p) · n = X · n

is formally soluble using the ideas of the Fredholm alternative.

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8. Integration on Manifolds

 8.5-4. Show that any symmetric two-tensor h on a compact Riemannian manifold (M, g) can be uniquely decomposed in the form h = £X g + k. where δk = 0, δ being the divergence of g, defined by δk = (£(·) g)∗ k, where (£(·) g)∗ is the adjoint of the operator X 7→ £X g. (See Berger and Ebin [1969] and Cantor [1981] for more information.)  8.5-5. Let α ∈ Ωk−1 (M ), β ∈ Ωk (M ), where M is a compact oriented Riemannian manifold with boundary. Z Show that hdα, βi − hα, δβi = α ∧ ∗β. ∂M (i) Hint: Show that ∗δβ = (−1)k d ∗ β and use Stokes theorem or Corollary 8.2.13. (ii) Now for α, β ∈ Ωk (M ), show that Z hdδα, βi − hδα, δβi =

δα ∧ ∗β Z∂M

hdα, dβi − hα, δdβi =

α ∧ ∗dβ ∂M

(iii) (Green’s formula) For α, β ∈ Ωk (M ), show that Z h∆α, βi − hα, ∆βi = (δα ∧ ∗β − δβ ∧ ∗α + α ∧ ∗dβ − β ∧ ∗dα). ∂M

Hint: Show first that Z h∆α, βi − hdα, dβi − hδα, δβi =

(δα ∧ ∗β − β ∧ ∗dα). ∂M

 8.5-6. (For students knowing algebraic topology.) Define relative cohomology groups and relate them to the Hodge decomposition for manifolds with boundary.  8.5-7.

Prove the local formulas

(δα)i1 ···ik ∂ 1 | det[grs ]|−1/2 gi1 r1 ···ik rk l = k+1 ∂x

k+1 X

! (−1)p g r1 j1 · · · g rp−1 jp−1 g ljp g rp jp+1 · · · g rk jk+1 βj1 ···jk+1 | det[grs ]|1/2

p=1

(δα)r1 ···rk ∂ 1 | det[gij ]|−1/2 l = k+1 ∂x

k+1 X

! (−1)p αr1 ···rp−1 lrp ···rk | det[gij ]|1/2

p=1

where i1 < · · · < ik and α ∈ Ωk+1 (M ) according to the following guidelines. First prove the second formula. Work in a chart (U, ϕ) with ϕ(U ) = B3 (0) = open ball of radius 3, and prove the formula on ϕ−1 (B1 (0)). For this, choose a function χ on Rn with supp(χ) ⊂ B3 (0) and χ|B1 (0) ≡ 1. Then extend χϕ∗ α to Rn , denote it by α0 and consider the set B4 (0). (i) Show from Exercise 8.5-5(i) that hdβ, α0 i = hβ, δα0 i for any β ∈ Ωk+1 (B4 (0)). (ii) In the explicit expression for hdβ, α0 i, perform an integration by parts and justify it. (iii) Find the expression for δα0 by comparing hβ, ∆α0 i with the expression found in (ii) and argue that it must hold on ϕ−1 (B1 (0)).

8.5 Introduction to Hodge–deRham Theory

527

 8.5-8. Let ϕ : M → M be a diffeomorphism of an oriented Riemannian manifold (M, g) and let δg denote the codifferential corresponding to the metric g and h , ig the inner product on Ωk (M ) corresponding to the metric g. Show that (i) hα, βig = hϕ∗ α, ϕ∗ βiϕ∗ g for α, β ∈ Ωk (M ) and (ii) δϕ∗ g (ϕ∗ α) = ϕ∗ (δg α) for α ∈ Ωk (M ). Hint: Use the fact that d and δ are adjoints.  8.5-9. (i) Let c1 and c2 be two differentiably homotopic curves and ω ∈ Ω1 (M ) a closed one-form. Show that Z Z ω= ω. c1

c2

(ii) Let M be simply connected. Show that H 1 (M ) = 0. Hint: For m0 ∈ M , let c be a curve from m0 to m ∈ M. Then f (m) = df = ω.

R c

ω is well defined by (i) and

(iii) Show that H 1 (S 1 ) 6= 0 by exhibiting a closed one-form that is not exact.  8.5-10. The Hopf degree theorem states that f and g : M n → S n are homotopic iff they have the same degree. By consulting references if necessary, prove this theorem in the context of Supplements 8.5A and B. Hint: Consult Guillemin and Pollack [1974] and Hirsch [1976].  8.5-11. What does the degree of a map have to do with Exercise 8.2-4 on integration over the fiber? Give some examples and a discussion.  8.5-12.

Show that the equations z 13 + sin(|z|2 )z 7 + 3z 4 + 2 = 0 z 8 + cos(|z|2 )z 5 + 5 log(|z|2 )z 4 + 53 = 0

have a root.  8.5-13. Let f : M → N where M and N are compact orientable boundaryless manifolds and N is contractible. Show that deg(f ) = 0. Conclude that the only contractible compact manifold (orientable or not) is the one-point space. Hint: Show that the oriented double covering of a contractible non-orientable manifold is contractible.  8.5-14. Show that every smooth map f : S n → Tn , n > 1 has degree zero. Conclude that S n and Tn are not diffeomorphic if n > 1. Hint: Show that f is homotopic to a constant map.

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8. Integration on Manifolds

Page 529

9 Applications

This version: March 1, 2007

This chapter presents some applications of manifold theory and tensor analysis to physics and engineering. Our selection is a of limited scope and depth, with the intention of providing an introduction to the techniques. There are many other applications of the ideas of this book as well. We list below a few selected references for further reading in the same spirit. 1. Arnol’d [1982], Abraham and Marsden [1978], Chernoff and Marsden [1974], Weinstein [1977], Marsden [1981], Marsden [1992], and Marsden and Ratiu [1999] for Hamiltonian mechanics. 2. Marsden and Hughes [1983] for elasticity theory. 3. Flanders [1963], von Westenholz [1981], and Bloch, Ballieul, Crouch, and Marsden [2003] for applications to control theory. 4. Hermann [1980], Knowles [1981], and Schutz [1980] for diverse applications. 5. Bleecker [1981] for Yang–Mills theory. 6. Misner, Thorne, and Wheeler [1973] and Hawking and Ellis [1973] for general relativity.

9.1

Hamiltonian Mechanics

Newton’s Second Law. Our starting point is Newton’s second law in R3 , which states that a particle which has mass m > 0, and is moving in a given potential field V (x) where x ∈ R3 , moves along a curve x(t) satisfying the equation of motion m¨ x = − grad V (x). If we introduce the momentum p = mx˙ and the energy 1 H(x, p) = kpk2 + V (x), 2m then the equation x˙ = p/m and Newton’s law become Hamilton’s equations: x˙ i =

∂H , ∂pi

p˙i = −

∂H , ∂xi

i = 1, 2, 3.

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9. Applications

To study this system of first-order equations for given H, we introduce the matrix   0 I J= , −I 0 where I is the 3 × 3 identity; note that the equations become ξ˙ = J grad H(ξ) where ξ = (x, p). In complex notation, setting z = x + ip, they may be written as z˙ = −2i

∂H . ∂ z¯

Suppose we make a change of coordinates, w = f (ξ), where f : R6 → R6 is smooth. If ξ(t) satisfies Hamilton’s equations, the equations satisfied by w(t) are w˙ = Aξ˙ = AJ gradξ H(ξ) = AJA∗ gradw H(ξ(w)), where Aij = (∂wi /∂ξ j ) is the Jacobian matrix of f , A∗ is the transpose of A and ξ(w) denotes the inverse function of f . The equations for w will be Hamiltonian with energy K(w) = H(ξ(w)) if AJA∗ = J. A transformation satisfying this condition is called canonical or symplectic. One of the things we do in this chapter is to give a coordinate free treatment of this and related concepts. The space R3 × R3 of the ξ’s is called the phase space. For a system of N particles one uses R3N × R3N . However, many fundamental physical systems have a phase space that is a manifold rather than Euclidean space, so doing mechanics solely in the context of Euclidean space is too constraining. For example, the phase space for the motion of a rigid body about a fixed point is the tangent bundle of the group SO(3) of 3 × 3 orthogonal matrices with determinant +1. This manifold is diffeomorphic to RP3 and is topologically nontrivial. To generalize the notion of a Hamiltonian system to the context of manifolds, we first need to geometrize the symplectic matrix J. In infinite dimensions a few technical points need attention before proceeding. Weak and Strong Metrics and Symplectic Forms. Let E be a Banach space and B : E × E → R a continuous bilinear mapping. Then B induces a continuous map B [ : E → E∗ , e 7→ B [ (e) defined by B [ (e) · f = B(e, f ). We call B weakly nondegenerate if B [ is injective, that is, B(e, f ) = 0 for all f ∈ E implies e = 0. We call B nondegenerate or strongly nondegenerate if B [ is an isomorphism. By the open mapping theorem, it follows that B is nondegenerate iff B is weakly nondegenerate and B [ is onto. If E is finite dimensional there is no difference between strong and weak nondegeneracy. However, if infinite dimensions the distinction is important to bear in mind, and the issue does come up in basic examples, as we shall see in Supplement 9.1A. Let M be a Banach manifold. By a weak Riemannian structure we mean a smooth assignment g : x 7→ h , ix = g(x) of a weakly nondegenerate inner product (not necessarily complete) to each tangent space Tx M. Here smooth means that in a local chart U ⊂ E, the mapping g : x 7→ h , ix ∈ L2 (E, E; R) is smooth, where L2 (E, E; R) denotes the Banach space of bilinear maps of E × E to R. Equivalently, smooth means g is smooth as a section of the vector bundle L2 (T M, T M ; R) whose fiber at x ∈ M is L2 (Tx M, Tx M ; R). By a Riemannian manifold we mean a weak Riemannian manifold in which h , ix is nondegenerate. Equivalently, the topology of h , ix is complete on Tx M, so that the model space E must be isomorphic to a Hilbert space. For example the L2 inner product Z 1 hhf, gii = f (x)g(x)dx on E = C 0 ([0, 1], R) 0

is a weak Riemannian metric on E but is not a Riemannian metric.

9.1 Hamiltonian Mechanics

531

9.1.1 Definition. Let P be a manifold modeled on a Banach space E. By a symplectic form we mean a two-form ω on P such that (i) ω is closed, that is, dω = 0; (ii) for each z ∈ P , ωz : Tz P × Tz P → R is weakly nondegenerate. If ωz in (ii) is nondegenerate, we speak of a strong symplectic form. If (ii) is dropped we refer to ω as a presymplectic form. (For the moment the reader may wish to assume P is finite dimensional, in which case the weak–strong distinction vanishes.) The Darboux Theorem. Our proof of this basic theorem follows a very elegant argument due to Moser [1965] (see also Weinstein [1969]). 9.1.2 Theorem (The Darboux Theorem). Let ω be a strong symplectic form on the Banach manifold P . Then for each x ∈ P there is a local coordinate chart about x in which ω is constant. Proof. The proof proceeds by the Lie transform method that is given in Theorem 6.4.7. We can assume P = E and x = 0 ∈ E. Let ω1 be the constant form equaling ω0 = ω(0). Let Ω = ω1 − ω and ωt = ω + tΩ, for 0 ≤ t ≤ 1. For each t, ωt (0) = ω(0) is nondegenerate. Hence by openness of the set of linear isomorphisms of E to E∗ , there is a neighborhood of 0 on which ωt is nondegenerate for all 0 ≤ t ≤ 1. We can assume that this neighborhood is a ball. Thus by the Poincar´e lemma, Ω = dα for some one-form α. We can suppose α(0) = 0. Define a smooth vector field Xt by iXt ωt = −α, which is possible since ωt is strongly non-degenerate. Since Xt (0) = 0, by Corollary 4.1.25, there is a sufficiently small ball on which the integral curves of Xt will be defined for time at least one. Let Ft be the flow of Xt starting at F0 = identity. By the Lie derivative formula for time-dependent vector fields (Theorem 6.4.4) we have d ∗ d (Ft ωt ) = Ft∗ (£Xt − ωt ) + Ft∗ ωt dt dt = Ft∗ diXt ωt + Ft∗ Ω = Ft∗ (d(−α) + Ω) = 0. Therefore, Ft∗ ω1 = F0∗ ω0 = ω, so F1 provides the chart transforming ω to the constant form ω1 .



We note without proof that such a result is not true for Riemannian structures unless they are flat. Also, the analogue of Darboux theorem is known to be not valid for weak symplectic forms. (For the example, see Abraham and Marsden [1978], Exercise 3.2-8 and for conditions under which it is valid, see Marsden [1981] and Bambusi [1999].) 9.1.3 Corollary.

If P is finite dimensional and ω is a symplectic form, then

(i) P is even dimensional, say dim P = 2n; (ii) locally about each point there are coordinates x1 , . . . , xn , y 1 , . . . , y n such that ω=

n X

dxi ∧ dy i .

i=1

Such coordinates are called canonical . Proof. By elementary linear algebra, any skew symmetric bilinear form that is nondegenerate has the canonical form   0 I , −I 0 where I is the n × n identity. (This is proved by the same method as Proposition 7.2.9.) This is the matrix version of (ii) pointwise on P . The result now follows from Darboux theorem. 

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9. Applications

As a bilinear form, ω is given in canonical coordinates by ω((x1 , y1 ), (x2 , y2 )) = hy2 , x1 i − hy1 , x2 i. In complex notation with z = x + iy it reads ω(z1 , z2 ) = − Imhz1 , z2 i. This form for canonical coordinates extends to infinite dimensions (see Cook [1966], Chernoff and Marsden [1974], and Abraham and Marsden [1978, Section 3.1] for details). Canonical Symplectic Forms. Of course in practice, symplectic forms do not come out of the blue, but must be constructed. The following constructions are basic results in this direction. 9.1.4 Definition. Let Q be a manifold modeled on a Banach space E. Let T ∗ Q be its cotangent bundle, and π : T ∗ Q → Q the projection. Define the canonical one-form θ on T ∗ Q by θ(α)w = α · T π(w), where α ∈

Tq∗ Q

∗

and w ∈ Tα (T Q). The canonical two-form is defined by ω = −dθ.

In a chart U ⊂ E, the formula for θ becomes θ(x, α) · (e, β) = α(e), ∗

∗

where (x, α) ∈ U × E and (e, β) ∈ E × E . If Q is finite dimensional, this formula may be written θ = pi dq i , where q 1 , . . . , q n , p1 , . . . , pn are coordinates for T ∗ Q and the summation convention is enforced. Using the local formula for d from formula (6) in the table of identities in §7.4, ω(x, α)((e1 , α1 ), (e2 , α2 )) = α2 (e1 ) − α1 (e2 ), or, in the finite-dimensional case, ω = dq i ∧ dpi . In the infinite-dimensional case one can check that ω is weakly nondegenerate and is strongly nondegenerate iff E is reflexive. If h , ix is a weak Riemannian (or pseudo-Riemannian) metric on Q, the smooth vector bundle map ϕ = g[ : T Q → T ∗ Q defined by ϕ(vx ) · wx = hvx , wx ix , x ∈ Q, is injective on fibers. If h , i is a strong Riemannian metric, then ϕ is a vector bundle isomorphism of T Q onto T ∗ Q. In any case, set Ω = ϕ∗ ω where ω is the canonical two–form on T ∗ Q. Clearly Ω is exact since Ω = −dΘ where Θ = ϕ∗ θ. In the finite-dimensional case, the formulas for Θ and Ω become Θ = gij q˙j dq i , and

∂gij j i q˙ dq ∧ dq k , ∂q k where q 1 , . . . , q n , q˙1 , . . . , q˙n are coordinates for T Q. This follows by substituting pi = gij q˙j into ω = dq i ∧dpi . In the infinite-dimensional case, if h , i is a weak metric, then ω is a weak symplectic form locally given by Ω = gij dq i ∧ dq˙j +

Θ(w, e)(e1 , e2 ) = −he, e1 ix , and Ω(x, e)((e1 , e2 ), (e3 , e4 )) = Dx he, e1 ix e3 − Dx he, e3 ix e1 + he4 , e1 ix − he2 , e3 ix , where Dx denotes the derivative with respect to x. One can also check that if h , ix is a strong metric and Q is modeled on a reflexive space, then Ω is a strong symplectic form.

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533

Symplectic Maps. Naturally, since we have the notion of a symplectic manifold, we should consider the mappings that preserve this structure. 9.1.5 Definition. Let (P, ω) be a symplectic manifold. A (smooth) map f : P → P is called canonical or symplectic when f ∗ ω = ω. It follows that f ∗ (ω ∧ · · · ∧ ω) = ω ∧ · · · ∧ ω (k times). If P is 2n-dimensional, then µ = ω ∧ · · · ∧ ω (n times) is nowhere vanishing, so is a volume form; for instance by a computation one finds µ to be a multiple of the standard Euclidean volume in canonical coordinates. In particular, note that symplectic manifolds are orientable. We call µ the phase volume or the Liouville form. Thus a symplectic map preserves the phase volume, and so is necessarily a local diffeomorphism. A map f : P1 → P2 between symplectic manifolds (P1 , ω1 ) and (P2 , ω2 ) is called symplectic if f ∗ ω2 = ω1 . As above, if P1 and P2 have the same dimension, then f is a local diffeomorphism and preserves the phase volume. Cotangent Lifts. bundle. 9.1.6 Proposition.

We now discuss symplectic maps induced by maps on the base space of a cotangent Let f : Q1 → Q2 be a diffeomorphism; define the cotangent lift of f by T ∗ f : T ∗ Q2 → T ∗ Q1 ;

T ∗ f (αq ) · v = αq · T f (v),

where q ∈ Q2 , αq ∈ Tq∗ Q2 and v ∈ Tf −1 (q) Q1 ; that is, T ∗ f is the pointwise adjoint of T f. Then T ∗ f is symplectic and in fact (T ∗ f )∗ θ1 = θ2 where θi is the canonical one-form on Qi , i = 1, 2. Proof. Let πi : T ∗ Qi → Qi be the cotangent bundle projection, i = 1, 2. For w in the tangent space to T ∗ Q2 at αq , we have (T ∗ f )∗ θ1 (αq )(w) = θ1 (T ∗ f (αq ))(T T ∗ f · w) = T ∗ f (αq ) · (T π1 · T T ∗ f · w) = T ∗ f (αq ) · (T (π1 ◦ T ∗ f ) · w) = αq · (T (f ◦ π1 ◦ T ∗ f ) · w = αq · (T π2 · w) = θ2 (αq ) · w since, by construction, f ◦ π1 ◦ T ∗ f = π2 .



In coordinates, if we write f (q 1 , . . . , q n ) = (Q1 , . . . , Qn ), then T ∗ f has the effect (q 1 , . . . , q n , p1 , . . . , pn ) 7→ (Q1 , . . . , Qn , P1 , . . . , Pn ), where pj =

∂Qi Pi ∂q j

(evaluated at the corresponding points). That this transformation is always canonical and in fact preserves the canonical one-form may be verified directly: Pi dQi = Pi

∂Qi k dq = pk dq k . ∂q k

Sometimes one refers to canonical transformations of this type as “point transformations” since they arise from general diffeomorphisms of Q1 to Q2 . Notice that lifts of diffeomorphisms satisfy f ◦ π2 = π1 ◦ T ∗ f ; that is, the following diagram commutes:

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9. Applications

T ∗ Q2

T ∗f

π2

- T ∗ Q1 π1 ? - Q1

? Q2 f Notice also that

T ∗ (f ◦ g) = T ∗ g ◦ T ∗ f and compare with T (f ◦ g) = T f ◦ T g. 9.1.7 Corollary. If Q1 and Q2 are Riemannian (or pseudo-Riemannian) manifolds and f : Q1 → Q2 is an isometry, then T f : T Q1 → T Q2 is symplectic, and in fact (T f )∗ Θ2 = Θ1 . Proof.

This follows from the identity T f = g2 # ◦ (T ∗ f )−1 ◦ g1 [ .

All maps in this composition are symplectic and thus T f is as well. Hamilton’s Equations. consider them.



So far no mention has been made of Hamilton’s equations. Now we are ready to

9.1.8 Definition. Let (P, ω) be a symplectic manifold. A vector field X : P → T P is called Hamiltonian if there is a C 1 function H : P → R such that iX ω = dH. We say X is locally Hamiltonian if iX ω is closed. We write X = XH because usually in examples one is given H and then one constructs the Hamiltonian vector field XH . If ω is only weakly nondegenerate, then given a smooth function H : P → R, XH need not exist on all of P . Rather than being a pathology, this is quite essential in infinite dimensions, for the vector fields then correspond to partial differential equations and are only densely defined. The condition iXH ω = dH is equivalent to ωz (XH (z), v) = dH(z) · v, for z ∈ P and v ∈ Tz P. Let us express this condition in canonical coordinates (q 1 , . . . , q n , p1 , . . . , pn ) on a 2n-dimensional symplectic manifold P , that is, when ω = dq i ∧ dpi . If X = Ai ∂/∂q i + B i ∂/∂pi , then iXH ω = iXH (dq i ∧ dqi ) = (iXH dq i )dpi − (iXH dpi )dq i = (Ai dpi − B i dq i ). This equals dH = iff Ai =

∂H ∂pi

∂H i ∂H dq + dqi ∂q i ∂pi and B i = −

∂H , ∂q i

9.1 Hamiltonian Mechanics

535

that is, XH

 n  X ∂H ∂ ∂H ∂ = − i . ∂pi ∂q i ∂q ∂pi i=1

If  J=

0 −I

 I , 0

where I is the n × n identity matrix, the formula for XH can be expressed as  XH =

∂H ∂H ,− i ∂pi ∂q

 = J grad H.

More intrinsically, one can write XH = ω # dH, so one sometimes says that XH is the symplectic gradient of H. Note that the formula XH = J grad H is a little misleading in this respect, since no metric structure is actually needed and it is really the differential and not the gradient that is essential. From the local expression for XH we see that (q i (t), pi (t)) is an integral curve of XH iff Hamilton’s equations hold; ∂H ∂H q˙i = , p˙i = − i . ∂pi ∂q Properties of Hamiltonian systems. The proofs of the following properties are a bit more technical for densely defined vector fields, so for purposes of these theorems, we work with C ∞ vector fields. 9.1.9 Theorem. Let XH be a Hamiltonian vector field on the (weak ) symplectic manifold (P, ω) and let Ft be the flow of XH . Then (i) Ft is symplectic, that is, Ft∗ ω = ω, and (ii) energy is conserved, that is, H ◦ Ft = H. Proof. (i) Since F0 = identity, it suffices to show that (d/dt)Ft∗ ω = 0. But by the basic connection between Lie derivatives and flows (§6.4 and §7.4): d ∗ F ω(x) = Ft∗ (£XH ω)(x) dt t = Ft∗ (diXH ω)(x) + Ft∗ (iXH dω)(x). The first term is zero because it is d2 H = 0 and the second is zero because ω is closed. (ii) By the chain rule, d (H ◦ Ft )(x) = dH(Ft (x)) · XH (Ft (x)) dt = ω(Ft (x))(XH (Ft (x)), XH (Ft (x))). But this is zero in view of the skew symmetry of ω.



A corollary of (i) in finite dimensions is Liouville’s theorem: Ft preserves the phase volume. This is seen directly in canonical coordinates by observing that XH is divergence-free.

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9. Applications

Poisson Brackets.

Define for any functions f, g : U → R, U open in P, their Poisson bracket by {f, g} = ω(Xf , Xg ).

Since £Xf g = iXf dg = iXf iXg ω = ω(Xg , Xf ) = −ω(Xf , Xg ) = −£Xg f, we see that {f, g} = £Xg f = −£Xf g. If ϕ : P1 → P2 , is a diffeomorphism where (P1 , ω1 ) and (P2 , ω2 ) are symplectic manifolds, then by the property ϕ∗ (£X α) = £ϕ∗ X ϕ∗ α of pull–back, we have ϕ∗ {f, g} = ϕ∗ (£Xf g) = £ϕ∗ Xf ϕ∗ g, and {ϕ∗ f, ϕ∗ g} = £Xϕ∗ f ϕ∗ g. Thus ϕ preserves the Poisson bracket of any two functions defined on some open set of P2 iff ϕ∗ Xf = Xϕ∗ f for all C ∞ functions f : U → Rwhere U is open in P2 . This says that ϕ preserves the Poisson bracket iff it preserves Hamilton’s equations. We have iXϕ∗ f ω = d(ϕ∗ f ) = ϕ∗ (df ) = ϕ∗ iXf ω = iϕ∗ Xf ϕ∗ ω, so that by the (weak) nondegeneracy of ω and the fact that any v ∈ Tz P equals some Xh (z) for a C ∞ function h defined in a neighborhood of z, we conclude that ϕ is symplectic iff ϕ∗ Xf = Xϕ∗ f for all C ∞ functions f : U → R, where U is open in P2 . We have thus proved the following. 9.1.10 Proposition. Let (P1 , ω1 ) and (P2 , ω2 ) be symplectic manifolds and ϕ : P1 → P2 a diffeomorphism. The following are equivalent: (i) ϕ is symplectic. (ii) ϕ preserves the Poisson bracket of any two locally defined functions. (iii) ϕ∗ Xf = Xϕ∗ f for any local f : U → R, where U is open in P2 (i.e., ϕ locally preserves Hamilton’s equations). Conservation of energy is generalized in the following way. 9.1.11 Corollary. (i) Let XH be a Hamiltonian vector field on the (weak ) symplectic manifold (P, ω) with (local ) flow Ft . Then for any C ∞ function f : U → R, U open in P , we have d (f ◦ Ft ) = {f, H} ◦ Ft = {f ◦ Ft , H}. dt (ii) The curve c(t) satisfies Hamilton’s equations defined by H if and only if d f (c(t)) = {f, H}(c(t)) dt for any C ∞ function f : U → R, where U is open in P .

9.1 Hamiltonian Mechanics

Proof.

537

(i) We compute as follows: d d (f ◦ Ft ) = Ft∗ f = Ft∗ £XH f = Ft∗ {f, H} = {Ft∗ f, H} dt dt

by the formula for Lie derivatives and the previous proposition. (ii) Since df (c(t))/dt = df (c(t)) · (dc/dt) and {f, H}(c(t)) = (£XH f )(c(t)) = df (c(t)) · XH (c(t)), the equation in the statement of the proposition holds iff c0 (t) = XH (c(t)) by the Hahn–Banach theorem and Corollary 4.2.14.  One writes f˙ = {f, H} to stand for the equation in (ii). This equation is called the equation of motion in Poisson bracket formulation. Two functions f, g : P → R are said to be in involution or to Poisson commute if {f, g} = 0. Any function Poisson commuting with the Hamiltonian of a mechanical system is, by Corollary 9.1.11, necessarily constant along on the flow of the Hamiltonian vector field. This is why such functions are called constants of the motion. A classical theorem of Liouville states that in a mechanical system with a 2n-dimensional phase space admitting k constants of the motion in involution and independent almost everywhere (i.e., the differentials are independent on an open dense set) one can reduce the dimension of the phase space to 2(n − k). In particular, if k = n, the equations of motion can be “explicitly” integrated. In fact, under certain additional hypotheses, the trajectories of the mechanical system are straight lines on high-dimensional cylinders or tori. If the motion takes place on tori, the explicit integration of the equations of motion goes under the name of finding action-angle variables. See Arnol’d [1982], and Abraham and Marsden [1978, pp. 392–400] for details and Exercise 9.1-4 for an example. In infinite-dimensional systems the situation is considerably more complicated. A famous example is the Korteweg–deVries (KdV) equation; for this example we also refer to Abraham and Marsden [1978, pp. 462–72] and references therein. The following supplement gives some elementary but still interesting examples of infinite-dimensional Hamiltonian systems.

Supplement 9.1A Two Infinite-Dimensional Examples 9.1.12 Example (The Wave Equation as a Hamiltonian System). The wave equation for a function u(x, t), where x ∈ Rn and t ∈ R is given by d2 u = ∇2 u + m2 u, dt2

(where m ≥ 0 is a constant),

with u and u˙ = ∂u/∂t given at t = 0. The energy is Z  Z 1 H(u, u) ˙ = |u| ˙ 2 dx + k∇uk2 dx . 2 We define H on pairs (u, u) ˙ of finite energy by setting P = H 1 (Rn ) × L2 (Rn ),

538

9. Applications

where H 1 consists of functions in L2 whose first (distributional) derivatives are also in L2 . (The Sobolev spaces H s defined this way are Hilbert spaces that arise in many problems involving partial differential equations. We only treat them informally here.) Let D = H 2 × H 1 and define XH : D → P by XH (u, u) ˙ = (u, ˙ ∇2 u + m2 u). Let the symplectic form be associated with the L2 metric as in the discussion following Definition 9.1.4, namely Z Z ω((u, u), ˙ (v, v)) ˙ = vu dx − uv ˙ dx. It is now an easy verification using integration by parts, to show that XH , ω and H are in the proper relation, so in this sense the wave equation is Hamiltonian. That the wave equation has a flow on P follows from (the real form of) Stone’s theorem (see Supplement 9.4A and Yosida [1995]).  9.1.13 Example (The Schr¨ odinger Equation). Let P = H a complex Hilbert space with ω = −2 Imh , i. Let Hop be a self-adjoint operator with domain D and let XH (ϕ) = iHop · ϕ and H(ϕ) = hHop ϕ, ϕi,

ϕ ∈ D.

Again it is easy to check that ω, XH and H are in the correct relation. Thus, XH is Hamiltonian. Note that ψ(t) is an integral curve of XH if 1 dψ = Hop ψ, i dt which is the abstract Schr¨ odinger equation of quantum mechanics. That XH has a flow is a special case of Stone’s theorem. We know from general principles that the flow eitHop will be symplectic. The additional structure needed for unitarity is exactly complex linearity.  Turning our attention to geodesics and to Lagrangian systems, let M be a (weak) Riemannian manifold with metric h , ix on the tangent space Tx M. The spray S : T M → T 2 M of the metric h , ix is the vector field on T M defined locally by S(x, v) = ((x, v), (v, γ(x, v))), for (x, v) ∈ Tx M, where γ is defined by hγ(x, v), wix ≡

1 Dx hv, vix · ω − Dx hv, wix · v 2

(9.1.1)

and Dx , hv, vix · w means the derivative of hv, v)x with respect to x in the direction of w. If M is finite dimensional, the Christoffel symbols are defined by putting γ i (x, v) = −Γijk (x)v j v k . Equation (9.1.1) is equivalent to 1 ∂gij i j k ∂gij i j k −Γijk v j v k wi = vv w − vw v ; 2 ∂xk ∂xk that is,   1 ∂ghk ∂gjh ∂gjk Γijk = g hi + − . 2 ∂xj ∂xk ∂xh The verification that S is well-defined independent of the charts is not too difficult. Notice that γ is quadratic in v. We will show below that S is the Hamiltonian vector field on T M associated with the kinetic energy hv, vi/2. The projection of the integral curves of S to M are called geodesics. Their local equations are thus x ¨ = γ(x, x), ˙ which in the finite-dimensional case becomes x ¨i + Γijk x˙ j x˙ k = 0,

i = 1, . . . , n.

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539

The definition of γ in equation (9.1.1) makes sense in the infinite as well as the finite-dimensional case, whereas the coordinate definition of Γijk makes sense only in finite dimensions. This provides a way to deal with geodesics in infinite-dimensional spaces. Let t 7→ (x(t), v(t)) be an integral curve of S. That is, x(t) ˙ = v(t)

and v(t) ˙ = γ(x(t), v(t)).

(9.1.2)

As we remarked, these will shortly be shown to be Hamilton’s equations of motion in the absence of a potential. To include a potential, let V : M → R be given. At each x, we have the differential of V , dV (x) ∈ Tx∗ M, and we define grad V (x) by hgrad V (x), wix = dV (x) · w.

(9.1.3)

(In infinite dimensions, it is an extra assumption that grad V exists, since the map Tx M → Tx∗ M induced by the metric is not necessarily bijective.) The equations of motion in the potential field V are given by x(t) ˙ = v(t);

v(t) ˙ = γ(x(t), v(t)) − grad V (x(t))).

(9.1.4)

The total energy, kinetic plus potential, is given by H(vx ) =

1 kvx k2 + V (x). 2

The vector field XH determined by H relative to the symplectic structure on T M induced by the metric, is given by equation (9.1.4). This will be part of a more general derivation of Lagrange’s equations given below.

Supplement 9.1B Geodesics Readers familiar with Riemannian geometry can reconcile the present approach to geodesics based on Hamiltonian mechanics to the standard one in the following way. Define the covariant derivative ∇ : X(M ) × X(M ) → X(M ) locally by (∇X Y )(x) = γx (X(x), Y (x)) + DY (x) · X(x), where X(x) and Y (x) are the local representatives of X and Y in the model space E of M and γx : E×E → E denotes the symmetric bilinear continuous mapping defined by polarization of the quadratic form γ(x, v). In finite dimensions, if E = Rn , then γ(x, v) is an Rn -valued quadratic form on Rn determined by the Christoffel symbols Γijk . The defining relation for ∇X Y becomes ∇X Y = X j Y k Γijk

k ∂ ∂ j ∂Y + X , i j ∂x ∂x ∂xk

where locally ∂ ∂ and Y = Y k k . i ∂x ∂x It is a straightforward exercise to show that the foregoing definition of ∇X Y is chart independent and that ∇ satisfies the following conditions defining an affine connection: X = Xi

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9. Applications

(i) ∇ is R-bilinear, (ii) for f : M → R smooth, ∇f X Y = f ∇ X Y

and ∇X f Y = f ∇x Y + X[f ]Y,

(iii) (∇X Y − ∇Y X)(x) = DY (x) · X(x) − DX(x) · Y (x) = [X, Y ](x), by the local formula for the Jacobi–Lie bracket of two vector fields. (The equivalence of sprays and affine connections was introduced by Ambrose, Palais, and Singer [1960].) If c(t) is a curve in M and X ∈ X(M ), the covariant derivative of X along c is defined by DX = ∇c X, dt where c is a vector field coinciding with c(t) ˙ at the points c(t). Locally, using the chain rule, this becomes d DX (c(t)) = −γc(t) (X(c(t)), X(c(t))) + X(c(t)), dt dt which also shows that the definition of DX/dt depends only on c(t) and not on how c˙ is extended to a vector field. In finite dimensions, the coordinate form of the preceding equation is  i d DX = Γijk (c(t))X j (c(t))c˙k (t) + X i (c(t)), dt dt where c(t) ˙ denotes the tangent vector to the curve at c(t). The vector field X is called autoparallel or is parallel-transported along c if DX/dt = 0. Thus c˙ is autoparallel along c iff in any coordinate system we have c¨(t) − γc(t) (c(t), ˙ c(t)) ˙ =0 or, in finite dimensions c¨i (t) + Γijk (c(t))c˙j (t)c˙k (t) = 0. That is, c˙ is autoparallel along c iff c is a geodesic. There is feedback between Hamiltonian systems and Riemannian geometry. For example, conservation of energy for geodesics is a direct consequence of their Hamiltonian character but can also be checked directly. Moreover, the fact that the flow of the geodesic spray on T M consists of canonical transformations is also useful in geometry, for example, in the study of closed geodesics (cf. Klingenberg [1978]). On the other hand, Riemannian geometry provides tools and concepts (such as parallel transport and curvature) that are useful in studying Hamiltonian systems.

Lagrangian Mechanics. We now generalize the idea of motion in a potential to that of a Lagrangian system; these are, however, still special types of Hamiltonian systems. We begin with a manifold M and a given function L : T M → R called the Lagrangian. In case of motion in a potential, take L(vx ) =

1 hvx , vx i − V (x), 2

which differs from the energy in that −V is used rather than +V. The Lagrangian L defines a map called the fiber derivative, FL : T M → T ∗ M as follows: let v, w ∈ Tx M, and set d FL(v) · w ≡ L(v + tw) . dt t=0

9.1 Hamiltonian Mechanics

541

That is, FL(v) · w is the derivative of L along the fiber in direction w. In the case of L(vx ) = (1/2)hvx , vx ix − V (x), we see that FL(vx ) · wx = hvx , wx ix , so we recover the usual map g [ : T M → T ∗ M associated with the bilinear form h , ix . Since T ∗ M carries a canonical symplectic form ω, we can use FL to obtain a closed two-form ωL on T M : ωL = (FL)∗ ω. A local coordinate computation yields the following local formula for ωL : if M is modeled on a linear space E, so locally T M looks like U × E where U ⊂ E is open, then ωL (u, e) for (u, e) ∈ U × E is the skew symmetric bilinear form on E × E given by ωL (u, e) · ((e1 , e2 ), (f1 , f2 )) = D1 (D2 L(u, e) · e1 ) · f1 − D1 (D2 L(u, e) · f1 ) · e1 + D2 (D2 L(u, e) · e1 ) · f2 − D2 (D2 L(u, e) · f1 ) · e2 ,

(9.1.5)

where D1 and D2 denote the indicated partial derivatives of L. In finite dimensions this reads ωL =

∂2L ∂2L i j dq ∧ dq + dq i ∧ dq˙j , ∂ q˙i ∂q j ∂ q˙i ∂ q˙j

where (q i , q˙j ) are the standard local coordinates on T Q. The two form ωL is (weakly) nondegenerate if D2 D2 L(u, e) is (weakly) nondegenerate; in this case L is called (weakly) nondegenerate. In the case of motion in a potential, nondegeneracy of ωL amounts to nondegeneracy of the metric h , ix . The action of L is defined by A : T M → R, A(v) = FL(v) · v, and the energy of L is E = A − L. In charts, E(u, e) = D2 L(u, e) · e − L(u, e), and in finite dimensions, E is given by the expression E(q, q) ˙ =

∂L i q˙ − L(q, q). ˙ ∂ q˙i

Given L, we say that a vector field Z on T M is a Lagrangian vector field or a Lagrangian system for L if the Lagrangian condition holds: ωL (v)(Z(v), w) = dE(v) · w

(9.1.6)

for all v ∈ Tq M and w ∈ Tv (T M ). Here dE denotes the differential of E. We shall see that for motion in a potential, this leads to the same equations of motion as we found before. If ωL were a weak symplectic form there would exist at most one such Z, which would be the Hamiltonian vector field for the Hamiltonian E. The dynamics is obtained by finding the integral curves of Z; that is, the curves t 7→ v(t) ∈ T M satisfying (dv/dt)(t) = Z(v(t)). From the Lagrangian condition it is easy to check that energy is conserved (even though L may be degenerate). 9.1.14 Proposition. Let Z be a Lagrangian vector field for L and let v(t) ∈ T M be an integral curve of Z. Then E(v(t)) is constant in t. Proof.

By the chain rule, d E(v(t)) = dE(v(t)) · v 0 (t) dt = dE(v(t)) · Z(v(t)) − ωL (v(t))(Z(v(t)), Z(v(t))) = 0

by skew symmetry of ωL .



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9. Applications

We now generalize our previous local expression for the spray of a metric, and the equations of motion in the presence of a potential. In the general case the equations are called Lagrange’s equations. 9.1.15 Proposition. Let Z be a Lagrangian system for L and suppose Z is a second-order equation (i.e., in a chart U × E for T M , Z(u, e) = (u, e, e, Z2 (u, e)) for some map Z2 : U × E → E). Then in the chart U × E, an integral curve (u(t), v(t)) ∈ U × E of Z satisfies Lagrange’s equations: that is, du (t) = v(t), dt

d (D2 L(u(t), v(t)) · w = D1 L(u(t), v(t)) · w dt

(9.1.7)

for all w ∈ E. If D2 D2 L, or equivalently ωL , is weakly nondegenerate, then Z is automatically second order. In case of motion in a potential, equation (9.1.7) reduces to the equations (9.1.4). Proof.

From the definition of the energy E we have locally DE(u, e) · (f1 , f2 ) = D1 (D2 L(u, e) · e) · f1 + D2 (D2 L(u, e) · e) · f2 − D1 L(u, e) · f1

(a term D2 L(u, e) · f2 has canceled). Locally we may write Z(u, e) = (u, e, Y1 (u, e), Y2 (u, e)). Using formula (9.1.5) for ωL , the condition on Z may be written D1 (D2 L(u, e) · Y1 (u, e)) · f1 − D1 (D2 L(u, e) · f1 ) · Y1 (u, e) + D2 (D2 L(u, e) · Y1 (u, e)) · f2 − D2 (D2 L(u, e) · f1 ) · Y2 (u, e) = D1 (D2 L(u, e) · e) · f1 − D1 L(u, e) · f1 + D2 (D2 L(u, e) · e) · f2 .

(9.1.8)

Thus, if ωL is a weak symplectic form, then D2 D2 L(u, e) is weakly nondegenerate, so setting f1 = 0 we get Y1 (u, e) = e, that is, Z is a second-order equation. In any case, if we assume that Z is second order, then condition (9.1.8) becomes D1 L(u, e) · f1 = D1 (D2 L(u, e) · f1 ) · e + D2 (D2 L(u, e) · f1 ) · Y2 (u, e) for all f1 ∈ E. If (u(t), v(t)) is an integral curve of Z and using dots to denote time differentiation, then u˙ = v and u ¨ = Y2 (u, u), ˙ so D1 L(u, u) ˙ · f1 = D1 (D2 L(u, u) ˙ · f1 ) · u˙ + D2 (D2 L(u, u) ˙ · f1 ) · u ¨ d ˙ · f1 = D2 L(u, u) dt by the chain rule.



The condition of being second order is intrinsic; Z is second order iff T τM ◦ Z = identity, where τM : T M → M is the projection. See Exercise 9.1-4. In finite dimensions Lagrange’s equations (9.1.7) take the form   dq i d ∂L ∂L i = q˙ , and = i , i = 1, . . . , n. i dt dt ∂ q˙ ∂q 9.1.16 Proposition. Assume ϕ : Q → Q is a diffeomorphism which leaves a weakly nondegenerate Lagrangian L invariant, that is, L ◦ T ϕ = L. Then c(t) is an integral curve of the Lagrangian vector field Z if and only if T ϕ ◦ c is also an integral curve.

9.1 Hamiltonian Mechanics

Proof.

543

Invariance of L under ϕ implies FL ◦ T ϕ = T ∗ ϕ−1 ◦ FL so that (T ϕ)∗ ωL = (FL ◦ T ϕ)∗ ω = (T ∗ ϕ−1 ◦ FL)∗ ω = (FL)∗ (T ∗ ϕ−1 )∗ ω = (FL)∗ ω = ωL

by Proposition 9.1.6. We also have for any v ∈ T Q, A(T ϕ(v)) = FL(T ϕ(v)) · T ϕ(v) = FL(v) and thus relation (9.1.6) implies dE = (T ϕ)∗ dE = (T ϕ)∗ iZ ωL = i(T ϕ)∗ Z (T ϕ)∗ ωL = i(T ϕ)∗ ZωL . Weak nondegeneracy of L yields then (T ϕ)∗ Z = Z which by Proposition 4.2.4 is equivalent to the statement in the proposition.  9.1.17 Example (Geodesics on the Poincar´e Upper Half Plane). Let  Q = (x, y) ∈ R2 | y > 0 so that T Q = Q × R2 . Define the Poincar´ e metric g on Q by g(x, y)((u1 , u2 ), (v 1 , v 2 )) =

u1 v 1 + u2 v 2 y2

and consider the Lagrangian (v 1 )2 + (v 2 )2 y2 defined by g. L is nondegenerate and thus by Proposition 9.1.15, the Lagrangian vector field Z defined by L is a second order equation. By local existence and uniqueness of integral curves, for every point (x0 , y0 ) ∈ Q and every vector (v01 , v02 ) ∈ T(x,y) Q, there is a unique geodesic γ(t) satisfying γ(0) = (x0 , y0 ), γ 0 (0) = (v01 , v02 ). We shall determine the geodesics of g by taking advantage of invariance properties of L. Note that the reflection r : (x, y) ∈ Q 7→ (−x, y) ∈ Q leaves L invariant. Furthermore, consider the homographies h(z) = (az + b)/(cz + d) for a, b, c, d ∈ R satisfying ad − bc = 1, where z = x + iy. Since   z Im[h(z)] = Im , (cz + d)2 L(x, y, v 1 , v 2 ) =

it follows that h(Q) = Q and since Tz h(v) = v/(cz + d)2 , where v = v 1 + iv 2 , it follows that h leaves L invariant. Therefore, by Proposition 9.1.16, γ is a geodesic if and only if r ◦ γ and h ◦ γ are. In particular, if γ(0) = (0, y0 ), γ 0 (0) = (0, u0 ), then (r ◦ γ)(0) = (0, y0 )

and

(r ◦ γ)0 (0) = (0, u0 ),

that is, γ = r ◦ γ and thus γ is the semiaxis y > 0, x = 0. Since for any q0 ∈ Q and tangent vector v0 to Q at q0 there exists a homography h such that h(iy0 ) = q0 , and the tangent of h at iy0 in the direction iu0 is v0 , it follows that the geodesics of g are images by h of the semiaxis { (0, y) | y > 0 }. If c = 0 or d = 0, this image equals the ray { (b/d, y) | y > 0 } or the ray { (a/c, y) | y > 0 }. If both c 6= 0, d 6= 0, then the image equals the arc of the circle centered at ((ad + bc)/2cd, 0) of radius 1/(2cd). Thus the geodesics of the Poincar´e upper half plane are either rays parallel to the y-axis or arcs of circles centered on the x-axis. (See Figure 9.1.1.) The Poincar´e upper half-plane is a model of the Lobatchevski geometry. Two geodesics in Q are called parallel if they do not intersect in Q. Given either a ray parallel to Oy or a semicircle centered on Ox and a point not on this geodesic, there are infinitely many semicircles passing through this point and not intersecting the geodesics, that is, through a point not on a geodesic there are infinitely many geodesics parallel to it. 

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9. Applications

y

x Figure 9.1.1. Geodesics in the Poincar´e upper half plane.

We close with a result for Lagrangian systems generalizing Example 4.1.23B. 9.1.18 Definition. A C 2 function V0 : [0, ∞] → R is called positively complete if it is decreasing and for any e > supi≥0 {V0 (t)}, satisfies Z ∞ [e − V0 (t)]−1/2 dt = +∞, where x ≥ 0. x

The last condition is independent of e. Examples of positively complete functions are −tα , −t[log(1 + t)]α , −t log(1 + t)[ log(log(1 + t) + 1)]α , etc. The following result is due to Marsden and Weinstein [1970]. 9.1.19 Theorem. Let Q be a complete weak Riemannian manifold, V : Q → R be a C 2 function and let Z be the Lagrangian vector field for 1 L(v) = kvk2 − V (τ (v)), 2 where τ : T Q → Q is the tangent bundle projection. Suppose there is a positively complete function v0 and a point q 0 ∈ Q such that V (q) ≥ V0 (d(q, q 0 )) for all q ∈ Q. Then Z is complete. Proof. Let c(t) be an integral curve of Z and let q(t) = (τ ◦ c)(t) be its projection in Q. Let q0 = q(0) and consider the differential equation on R dV0 f 00 (t) = − (f (t)) (9.1.9) df p with initial conditions f (0) = d(q 0 , q0 ), f 0 (0) = 2(β − V0 (f (0))), where β = E(c(t)) = E(c(0)) ≥ V (q0 ). We can assume β > V (q0 ), for if β = V (q0 ), then q(0) ˙ = 0; now if q(t) ˙ = 0, the conclusion is trivially satisfied, so we need to work under the assumption that there exists a t0 for which q(t ˙ 0 ) 6= 0; by time translation we can assume t0 = 0. We show that the solution f (t) of equation (9.1.9) is defined for all t ≥ 0. Multiplying both sides of equation (9.1.9) by f 0 (t) and integrating yields Z s 1 0 2 f (t) = β − V0 (f (t)), that is, t(s) = [2(β − V0 (u))]−1/2 du. 2 d(q 0 ,q0 ) By hypothesis, the integral on the right diverges and hence t(s) → +∞ as s → +∞. This shows that f (t) exists for all t ≥ 0. For t ≥ 0, conservation of energy and the estimate on the potential V imply Z t d(q(t), q 0 ) ≤ d(q(t), q0 ) + d(q0 , q 0 ) ≤ d(q 0 , q0 ) + kq(s)kds ˙ 0 Z t = d(q 0 , q0 ) + [2(β − V0 (q(s))]1/2 ds 0 Z t ≤ d(q 0 , q0 ) + [2(β − V0 (d(q(s)), q 0 ))]1/2 ds. 0

9.1 Hamiltonian Mechanics

545

Since f (t) = d(q 0 , q0 ) +

Z

1

[2(β − V0 (f (s))]1/2 ds,

0 0

it follows that d(q(t), q ) ≤ f (t); see Exercise 4.1-9(v) or the reasoning in Example 4.1.23B plus an approximation of d(q(t), q 0 ) by C 1 functions. Hence if Q is finite dimensional, q(t) remains in a compact set for finite t-intervals, t ≥ 0. Therefore c(t) does as well, V (q(t)) being bounded below on such a finite t-interval. Proposition 4.1.19 implies that c(t) exists for all t ≥ 0. The proof in infinite dimensions is done in Supplement 9.1C. If Ft is the local flow of Z, from τ (F−t (v)) = τ (Ft (−v)) (reversibility), it follows that c(t) exists also for all t ≤ 0 and so the theorem is proved. 

Supplement 9.1C Completeness of Lagrangian Vector Fields on Hilbert Manifolds This supplement provides the proof of Theorem 9.1.19 for infinite dimensional Riemannian manifolds. We start with a few facts of general interest. Let (Q, g) be a Riemannian manifold and τ : T Q → Q the tangent bundle projection. For v ∈ Tq Q, the subspace Vv = ker Tv τ = Tv (Tq Q) ⊂ Tv (T Q) is called the vertical subspace of Tv (T Q). The local expression of the covariant derivative ∇ defined by g in Supplement 9.1B shows that ∇Y X depends only on the point values of Y and thus it defines a linear map (∇X)(q) : v ∈ Tq Q 7→ (∇Y X)(q) ∈ Tq Q, where Y ∈ X(Q) is any vector field satisfying Y (q) = v. Let jv : Tq Q → Tv (Tq Q) = Vv denote the isomorphism identifying the tangent space to a linear space with the linear space itself and consider the map jv ◦ (∇X)(q) : Tq Q → Vv . Define the horizontal map hv : Tq Q → Tv (T Q) by hv = Tq X − jv ◦ (∇X)(q) where v ∈ Tq Q and X ∈ X(Q) satisfies X(q) = v. Locally, if E is the model of Q, hv has the expression hv : (x, u) ∈ U × E 7→ (x, v, u − γx (u, v)) ∈ U × E × E × E. This shows that hv is a linear continuous injective map with split image. The image of hv is called the horizontal subspace of Tv (T Q) and is denoted by Hv . It is straightforward to check that Tv (T Q) = Vv ⊕Hv and that Tv τ |Hv : Hv → Tq Q, jv : Tq Q → Vv are Banach space isomorphisms. Declaring them to be isometries and Hv perpendicular to Vv gives a metric g T on T Q. We have proved that if (Q, q) is a (weak ) Riemannian manifold, then g induces a metric g T on T Q1 . The following result is taken from Ebin [1970]. 9.1.20 Proposition. 1 Sometimes

If (Q, g) is a complete (weak ) Riemannian manifold then so is (T Q, g T ).

this metric is called the Sasaki metric.

546

9. Applications

Proof. Let {vn } be a Cauchy sequence in T Q and let qn = τ (vn ). Since τ is distance decreasing it follows that {qn } is a Cauchy sequence in Q and therefore convergent to q ∈ Q by completeness of Q. If E is the model of Q, E is a Hilbert space, again by completeness of Q. Let (U, ϕ) be a chart at q and assume that U is a closed ball in the metric defined by g of radius 3. Also, assume that Tv T ϕ : Tv (T M ) ∈ E×E is an isometry for all v → Tq Q which implies that for  small enough there is a C > 0 such that kT T ϕ(w)k ≤ Ckwk for all w ∈ T T U. This means that all curves in T U are stretched by T ϕ by a factor of C. Let V ⊂ U be the closed ball of radius  centered at q and let n,m be large enough so the distance between vn and vm is smaller than  and vn , vm ∈ T V. If γ is a path from vn to vm of length < 2, then τ ◦ γ is a path from qn to qm of length < 2 and therefore τ ◦ γ ⊂ U, which in turn implies that γ ⊂ T U. Moreover, T ϕ ◦ γ has length < 2C and therefore the distance between T ϕ(vn ) and T ϕ(vm ) in E is at most 2C. This shows that {T ϕ(vn )} is a Cauchy sequence in E and hence convergent. Since T ϕ is a diffeomorphism, {vn } is convergent.  In general, completeness of a vector field on M implies completeness of the first variation equation on T M. Proof of Theorem 9.1.19 in infinite dimensions. Let c : ]a, b[ → T Q be a maximal integral curve of Z. We shall prove that limt↑b c(t) exists in T Q which implies, by local existence and uniqueness, that c can be continued beyond b, that is, that b = +∞. One argues similarly for a. We have shown that q(t) = (τ ◦c)(t) is bounded on finite t-intervals. Since V (q(t)) is bounded on such a finite t-interval, it follows that q(t) ˙ = c(t) is bounded in the metric defined by g T on T Q. By the mean value inequality it follows that if tn ↑ b, then {q(tn )} is a Cauchy sequence and therefore convergent since Q is complete. Next we show by the same argument that if tn ↑ b, then {c(tn )} is Cauchy, that is, we will show that c(t) ˙ is bounded on bounded t-intervals. Write c(t) ˙ = Z(c(t)) = S(c(t)) + V (c(t)), where S is the spray of g and represents the horizontal part of Z and V is the vertical part of Z. Since V (c(t)) depends only on q(t) and since q(t) extends continuously to q(b), it follows that kV (c(t))k is bounded as t ↑ b. Since 2 kS(c(t))k = kc(t)k and kc(t)k ˙ = kS(c(t))k2 + kV (c(t))k2

by the definition of the metric g T , it follows that kc(t)k ˙ remains bounded on finite t-intervals. Therefore {c(tn )} is Cauchy and Proposition 9.1.20 implies that c(t) can be continuously extended to c(b).  Remark. Note that completeness of Q, an estimate on the potential V , and conservation of the energy E, replaces “compactness” in Proposition 4.1.19 with “boundedness.” 

Exercises  9.1-1. Let (M, ω) be a symplectic manifold with ω = dθ and f : M → M a local diffeomorphism. Prove that f is a symplectic iff for every compact oriented two-manifold B with boundary, B ⊂ M, we have Z Z θ= θ. ∂B

f (∂B)

 9.1-2 (J. Moser). Use the method of proof of Darboux theorem to prove that if M is a compact manifold, µ and ν are two volume forms with the same orientation, and Z Z µ = ν, then there is a diffeomorphism f : M → M such that f ∗ ν = µ.

9.1 Hamiltonian Mechanics

Hint: Use the Lie transform method. Since Z Z µ = ν,

547

µ − ν = dα

(see Supplement 7.5B); put νt = tν + (1 − t)µ and define Xt by letting the interior product of Xt with νt be α. Let ϕt be the flow of Xt and set f = ϕ1 .  9.1-3.

On T ∗ R3 , consider the periodic three-dimensional Toda lattice Hamiltonian, H(q, p) =

1 kpk2 + eq1 −q2 + eq2 −q3 + eq3 −q1 . 2

(i) Write down Hamilton’s equations. (ii) Show that f1 (q, p) = p1 + p2 + p3 , f2 = H, and 1 f3 (q, p) = (p31 + p32 + p33 ) + p1 (exp(q 1 − q 2 ) + exp(q 3 − q1 )) 3 + p2 (exp(q 1 − q 2 ) + exp(q 2 − q 3 )) + p3 (exp(q 1 − q 2 ) + exp(q 2 − q 3 )) are in involution and are independent everywhere. (iii) Prove the same thing for g1 = f1 ,

g2 (q, p) = exp(q 1 − q 2 ) + exp(q 2 − q 3 ) + exp(q 3 − q 1 ),

and g3 (q, p) = p1 p2 p3 − p1 exp(q 2 − q 3 ) − p2 exp(q 3 − q 1 ) − p3 exp(q 1 − q 2 ). (iv) Can you establish (iii) without explicitly computing the Poisson brackets? Hint: Express g1 , g2 , g3 as polynomials of f1 , f2 , f3 .  9.1-4. A second-order equation on a manifold M is a vector field X on T M such that TτM ◦X = IdT M . Show that (i) X is a second-order equation iff for all integral curves c of X in T M we have (τM ◦ c)0 = c. One calls τM ◦ c a base integral curve. (ii) X is a second-order equation iff in every chart the local representative of X has the form (u, e) 7→ (u, e, e, V (u, e)). (iii) If M is finite dimensional and X is a second-order equation, then the base integral curves satisfy d2 x(t) = V (x(t), x(t)), ˙ dt2 where (x, x) ˙ denotes standard coordinates on T M .  9.1-5 (Noether theorem).

Prove the following result for Lagrangian systems.

548

9. Applications

Theorem. Let Z be a Lagrangian vector field for L : T M → R and suppose Z is a second-order equation. Let Φt be a one-parameter group of diffeomorphisms of M generated by the vector field Y : M → T M. Suppose that for each real number t, L◦T Φt = L. Then the function P (Y ) : T M → R, defined by P (Y )(v) = FL(v)·Y is constant along integral curves of Z .  9.1-6. Use Exercise 9.1-5 to show conservation of linear (resp., angular) momentum for the motion of a particle in R3 moving in a potential that has a translation (resp., rotational) symmetry.  9.1-7.

Consider R2n+2 with coordinates (q 1 , . . . , q n , E, p1 , . . . , pn , t) and define the symplectic form ω = dq i ∧ dpi + dE ∧ dt.

Consider the function P (q, p, E, t) = H(q, p, t) − E. Show that the vector field X = q˙i

∂ ∂ ∂ ∂ + p˙i + E˙ + t˙ ∂q i ∂pi ∂t ∂E

˙ defined by iX ω = dP reproduces familiar equations for q, ˙ p, ˙ t˙ and E.  9.1-8.

Show that the wave equation (see Supplement 9.1A) may be derived as a Lagrangian system.

 9.1-9. Refer to Example 9.1.13 on the Schr¨odinger equation. Let A and B be self adjoint operators on H and let fA : H → R be given by fA (ψ) = hψ, Aψi (the expectation value of A in the state ψ). Show that Poisson brackets and commutators are related by fi[A,B] = {fA , fB }.  9.1-10. Show that the geodesic flow of a compact Riemannian manifold is complete. (Warning: Compact pseudo-Riemannian manifolds need not be complete; see Wolf [1967] and Marsden [1973].)  9.1-11. Show that any isometry of a weak pseudo-Riemannian manifold maps geodesics to geodesics. (A map ϕ : Q → Q is called an isometry if ϕ∗ g = g, where g is the weak pseudo-Riemannian metric on Q.)  9.1-12.

Let (Q, g) be a weak Riemannian manifold.

(i) If Ft is the flow of the spray of g show that τ (Ft (sv)) = τ (Fst (v)), where τ : T Q → Q is the projection. (ii) Let U be any bounded set in Tq Q. Show that there is an  > 0 such that for any v ∈ U, the integral curve of the spray with initial condition v exists for a time ≥ . Hint: Let V ⊂ U be an open neighborhood of v such that all integral curves starting in V exist for time ≥ δ. Find R > 0 such that R−1 U ⊂ V and use (i).  9.1-13. A weak pseudo-Riemannian manifold (Q, g) is called homogeneous if for any x, y ∈ Q there is an isometry ϕ such that ϕ(x) = y. Show that homogeneous weak Riemannian manifolds are complete by using Exercises 9.1-11 and 9.1-12. Hint: Put the initial condition v in a ball B and choose  as in Exercise 9.1-12(ii). Let v(t) be the integral curve of S through v and let q(t) be the corresponding geodesic. The geodesic starting at q() in the direction v(), is ϕ applied to the geodesic through q = τ (v) in the direction T ϕ−1 (v)(); ϕ is the isometry sending q to q(). The latter geodesic lies in the ball B, so it exists for time ≥ .

9.2

Fluid Mechanics

We present a few of the basic ideas concerning the motion of an ideal fluid from the point of view of manifolds and differential forms. This is usually done in the context of Euclidean space using vector calculus. For the latter approach and additional details, the reader should consult one of the standard texts on the subject

9.2 Fluid Mechanics

549

such as Batchelor [1967], Chorin and Marsden [1993], or Gurtin [1981]. The use of manifolds and differential forms can give additional geometric insight. The present section is, for expository reasons, somewhat superficial and is intended only to indicate how to use differential forms and Lie derivatives in fluid mechanics. Once the basics are understood, more sophisticated questions can be asked, such as: in what sense is fluid mechanics an infinite-dimensional Lagrangian or Hamiltonian system? For the answer, see Arnol’d [1982], Abraham and Marsden [1978], Marsden and Weinstein [1983], Marsden, Ratiu, and Weinstein [1984] and Holm, Marsden, and Ratiu [1998]. For analogous topics in elasticity, see Marsden and Hughes [1983], and for plasmas, see §9.4 and Marsden and Weinstein [1982]. Let M be a compact, oriented finite-dimensional Riemannian n-manifold, possibly with boundary. Let the Riemannian volume form be denoted µ ∈ Ωn (M ), and the corresponding volume element dµ. Usually M is a bounded region with smooth boundary in two- or three-dimensional Euclidean space, oriented by the standard basis, and with the standard Euclidean volume form and inner product. Imagine M to be filled with fluid and the fluid to be in motion. Our object is to describe this motion. Let x ∈ M be a point in M and consider the particle of fluid moving through x at time t = 0. For example, we can imagine a particle of dust suspended in the fluid; this particle traverses a trajectory which we denote ϕt (x) = ϕ(x, t). Let u(x, t) denote the velocity of the particle of fluid moving through x at time t. Thus, for each fixed time, u is a vector field on M . See Figure 9.2.1. We call u the velocity field of the fluid . Thus the relationship between u and ϕt is dϕt (x) = u(ϕt (x), t); dt that is, u is a time-dependent vector field with evolution operator ϕt in the same sense as was used in §4.1. For each time t, we shall assume that the fluid has a well-defined mass density and we write ρt (x) = ρ(x, t). Thus if W is any subregion of M, we assume that the mass of fluid in W at time t is given by Z m(W, t) = ρt dµ. W

Our derivation of the equations is based on three basic principles, which we shall treat in turn: 1. Mass is neither created nor destroyed. 2. (Newton’s second law) The rate of change of momentum of a portion of the fluid equals the force applied to it. 3. Energy is neither created nor destroyed.

M x

trajectory of fluid particle moving through

Figure 9.2.1. The trajectory of a fluid particle

550

9. Applications

1. Conservation of mass. This principle says that the total mass of the fluid, which at time t = 0 occupied a nice region W , remains unchanged after time t; that is, Z Z ρt dµ = ρ0 dµ. ϕt (W )

W

(We call a region W “nice” when it is an open subset of M with smooth enough boundary to allow us to use Stokes’ theorem.) Let us recall how to use the transport theorem 8.1.12 to derive the continuity equation. Using the change-of-variables formula, conservation of mass may be rewritten as Z Z ∗ ϕt (ρt µ) = ρ0 µ W

W

for any nice region W in M , which is equivalent to ϕ∗t (ρt µ) = ρ0 µ,

or

(ϕ∗t ρt )J(ϕt ) = ρ0 ,

where J(ϕt ) is the Jacobian of ϕt . This in turn is equivalent to   ∂ρ d µ 0 = ϕ∗t (ρt , µ) = ϕ∗t £u (ρt , µ) + dt ∂t    ∂ρ = ϕ∗t u[ρt ] + ρt div u + µ ∂t    ∂ρ = ϕ∗t div(ρt u) + µ ∂t by the Lie derivative formula and Proposition 7.5.17. Thus ∂ρ + div(ρt u) = 0 ∂t is the differential form of the law of conservation of mass, also known as the continuity equation. Because of shock waves that could be present, ρ and u may not be smooth enough to justify the steps leading to the differential form of this law; the integral form will then be the one to use. Also note that the Riemannian metric has as yet played no role; only the volume element of M was needed. 2. Balance of momentum. Newton’s second law asserts that the rate of change of momentum of a portion of the fluid equals the total force applied to it. To see how to apply this principle on a general manifold, let us discuss the situation M ⊂ R3 first. Here we follow the standard vector calculus conventions and write the velocity fields in boldface type. The momentum of a portion of the fluid at time t that at time t = 0 occupied the region W is Z ρu dµ. ϕt (W ) 3

Here and in what follows the integral is R -valued, so we apply all theorems on integration componentwise. For any continuum, forces acting on a piece of material are of two types. First there are forces of stress, whereby the piece of material is acted on by forces across its surface by the rest of the continuum. Second, there are external, or body forces, such as gravity or a magnetic field, which exert a force per unit volume on the continuum. The clear formulation of surface stress forces in a continuum is usually attributed to Cauchy. We shall assume R that the body forces are given by a given force density b, that is, the total body forces acting on W are ρb dµ. In continuum mechanics the forces of stress are assumed to be of the form W R σ(x, t) · n da, where da is the induced volume element on the boundary, n is the outward unit normal, ∂W and σ(x, t) is a time-dependent contravariant symmetric two-tensor, called the Cauchy stress tensor . The contraction σ(t, x) · n is understood in the following way: if σ has components σ ij and n has components nk ,

9.2 Fluid Mechanics

551

then σ · n is a vector with components (σ · n)i = gjk σ ij nk , where g is the metric (in our case gjk = δjk ). The vector σ · n, called the Cauchy traction vector , measures the force of contact (per unit area orthogonal to n) between two parts of the continuum. (A theorem of Cauchy states that if one postulates the existence of a continuous Cauchy traction vector field T(x, t, n) satisfying balance of momentum, then it must be of the form σ · n, for a two-tensor, σ; moreover if balance of moment of momentum holds, σ must be symmetric. See Chorin and Marsden [1993], Gurtin [1981], or Marsden and Hughes [1983] for details.) Balance of momentum is said to hold when Z Z Z d ρu dµ = ρb dµ + σ · n da dt ϕt (W ) ϕt (W ) ∂ϕt (W ) for any nice region W in M = R3 . If div σ denotes the vector with components (div(σ 1i ), div(σ 2i ), div(σ 3i )), then by Gauss’ theorem Z Z σ · n da = ∂ϕt (W )

(div σ)dµ. ϕt (W )

By the change-of-variables formula and Lie derivative formula, we get Z Z d ∗ d i ρu dµ = ϕt (ρui dµ) dt ϕt (W ) dt W   Z ∂(ρui ) = + (£u ρ)ui + ρ£u ui + ρui div u dµ, ∂t ϕt (W ) so that the balance of momentum is equivalent to ∂ρ i ∂ui u +ρ + (dρ · u)ui + ρ£u ui + ρui div u = ρbi + (div σ)i . ∂t ∂t But dρ·u+ρ div u = div(ρu) and by conservation of mass, ∂ρ/∂t+div(ρu) = 0. Also, £u ui = (∂ui /∂xj )uj = (u · ∇)ui , so we get ∂u 1 + (u · ∇)u = b + div σ, ∂t ρ which represents the basic equations of motion. Here the quantity ∂u/∂t +(u · ∇)u is usually called the material derivative and is denoted by Du/dt. These equations are for any continuum, be it elastic or fluid. An ideal fluid is by definition a fluid whose Cauchy stress tensor σ is given in terms of a function p(x, t) called the pressure, by σ ij = −pg ij . In this case, balance of momentum in differential form becomes the Euler equations for an ideal fluid : ∂u 1 + (u · ∇)u = b − grad p. ∂t ρ The assumption on the stress σ in an ideal fluid means that if S is any fluid surface in M with outward unit normal n, then the force of stress per unit area exerted across a surface element S at x with normal n at time t is −p(x, t)n (see Figure 9.2.2). Let us return to the context of a Riemannian manifold M . First, it is not clear what the vector-valued integrals should mean. But even if we could make sense out of this, using, say parallel transport, there is a more serious problem with the integral form of balance of momentum as stated. Namely, if one changes coordinates, then balance of momentum does not look the same. One says that the integral form of balance of momentum is not covariant. Therefore we shall concentrate on the differential form and from now on we shall deal only with ideal fluids. (For a detailed discussion of how to formulate the basic integral balance laws of continuum mechanics covariantly, see Marsden and Hughes [1983]. A genuine difficulty with shock wave theory is that the notion of weak solution is not a coordinate independent concept.)

552

9. Applications

S

Figure 9.2.2. The stress in an ideal fluid is given by the pressure

Rewrite Euler’s equations in R3 with indices down; that is, take the flat of these equations. Then the i-th equation, i = 1, 2, 3 is ∂ui ∂ui ∂ui ∂ui 1 ∂p + u 1 1 + u 2 2 + u 3 3 = bi − . ∂t ∂x ∂x ∂x ρ ∂xi We seek an invariant meaning for the sum of the last three terms on the left-hand side. For fixed i this expression is   ∂ui ∂ui ∂uj ∂uj 1 b 2 uj j = uj j + uj i − uj i = (£u u )i − dkuk . ∂x ∂x ∂x ∂x 2 i That is, Euler’s equations can be written in the invariant form 1 ∂u[ 1 + £u u[ − d(u[ (u)) = − dp + b[ . ∂t 2 ρ We postulate this equation as the balance of momentum in M for an ideal fluid. The reader familiar with Riemannian connections (see Supplement 9.1B) can prove that this form is equivalent to the form ∂u 1 + ∇u u = − grad p + b ∂t ρ by showing that 1 £u u[ = (∇u u)[ + d(u[ (u)). 2 where ∇u u is the covariant derivative of u along itself, with ∇ the Riemannian connection given by g. The boundary conditions that should be imposed come from the physical significance of ideal fluid: namely, no friction should exist between the fluid and ∂M ; that is, u is tangent to ∂M at points of ∂M. Summarizing, the equations of motion of an ideal fluid on a compact Riemannian manifold M with smooth boundary ∂M and outward unit normal n are ∂u[ 1 1 + £u u[ − d(u[ (u)) = − dp + b[ ∂t 2 ρ

and

∂ρ + div(ρu) = 0. ∂t

9.2 Fluid Mechanics

553

We also have the boundary conditions uk∂M,

that is, u · n = on ∂M ;

and initial conditions u(x, 0) = u0 (x)

given on M.

We shall assume b = 0 from now on for simplicity. 3. Conservation of energy. A basic problem of ideal fluid dynamics is to solve the initial-boundaryvalue problem. The unknowns are u, ρ, and p, that is, n + 2 scalar unknowns. We have, however, only n + 1 equations. Thus one might suspect that to specify the fluid motion, one more equation is needed. This is in fact true and the law of conservation of energy will supply the necessary extra equation in fluid mechanics. (The situation is similar for general continua; see Marsden and Hughes [1983].) For a fluid moving in M with velocity field u, the kinetic energy of the fluid is Z 1 ρkuk2 dµ Ekinetic = 2 M where kuk2 = hu, ui is the square length of the vector function u. We assume that the total energy of the fluid can be written Etotal = Ekinetic + Einternal , where Einternal is the energy that relates to energy we cannot “see” on a macroscopic scale and derives from sources such as intermolecular potentials and molecular vibrations. If energy is pumped into the fluid or if we allow the fluid to do work, Etotal will change. We describe two particular examples of energy equations that are useful. A.

Assume that Einternal = constant. Then we ought to have Ekinetic as a constant of the motion; that is, Z d 1 ( ρkuk2 dµ) = 0. dt 2 M

To deal with this equation it is convenient to use the following. 9.2.1 Theorem (Transport Theorem with Mass Density). Let f be a time-dependent smooth function on M . Then if W is any (nice) open set in M , Z Z Df d ρf dµ = ρ dµ, dt ϕt (W ) dt ϕt (W ) where Df /dt = ∂f /∂t + £u f . Proof. By the change of variables formula, the Lie derivative formula, div(ρu) = u[ρ] + ρ div(u), and conservation of mass, we have   Z Z Z d d ∂(ρf ) ρf dµ = ϕ∗t (ρf µ) = ϕ∗t µ + £u (ρf µ) dt ϕt (W ) dt W ∂t W   Z ∂f ∂ρ = f µ + ρ µ + u[ρ]f µ + ρ(£u f )µ + ρ£u µ ∂t ∂t ϕt (W )      Z ∂ρ ∂f = + u[ρ] + ρ div u f µ + ρ + £u f µ ∂t ∂t ϕt (W )      Z ∂ρ ∂f = f + div(ρu) + ρ + £u f µ ∂t ∂t ϕt (W )   Z ∂f = ρ + £u f µ.  ∂t ϕt (W )

554

9. Applications

Making use of      
£u kuk2 = £u u[ (u) = £u u[ (u) = d u[ (u) (u), the transport lemma, and Euler’s equations, we get  Z    Z 1 ∂kuk2 d 1 ρkuk2 dµ = ρ + £u kuk2 dµ 0= dt 2 M 2 M ∂t Z Z [ ∂u 1 = ρ · u dµ + (£u u[ ) · u dµ ∂t 2 M M Z Z Z ∂u[ 1 [ · u dµ + ρd(u[ (u)) · u dµ = ρ(£u u ) · u dµ − ρ ∂t 2 M M M Z Z =− dp · u dµ = {(div u)pµ − £u (pµ)} (by the Leibniz rule for £u ) M M Z Z (div u)pµ. {(div u)pµ − d(iu pµ)} = = M

M

The last equality is obtained by Stokes’ theorem and the boundary conditions 0 = (u · n)da = iu µ. If we imagine this to hold for the same fluid in all conceivable motions, we are forced to postulate one of the additional equations div u = 0 or p = 0. The case div u = 0 is that of an incompressible fluid . Thus in this case the Euler equations are ∂u[ 1 1 + £u u[ − dkuk2 = − dp ∂t 2 ρ ∂ρ + div(ρu) = 0 ∂t div u = 0 with the boundary condition iu µ = 0 on ∂M and initial condition u(x, 0) = u0 (x). The case p = 0 is also possible but is less interesting. For a homogeneous incompressible fluid, with constant density ρ, Euler’s equations can be reformulated in terms of the Hodge decomposition theorem (see §7.5). Nonhomogeneous incompressible flow requires a weighted Hodge decomposition (see Marsden [1976]). Recall that any one–form α can be written in a unique way as α = dβ + γ, where δγ = 0. Define the linear operator  P : Ω1 (M ) → γ ∈ Ω1 (M ) | δγ = 0 by P(α) = γ. We are now in a position to reformulate Euler’s equations. Let Ω1δ=0 be the set of C ∞ one-forms γ with δγ = 0 and γ tangent to ∂M ; that is, ∗γ|∂M = 0. Let T : Ω1δ=0 → Ω1δ=0 be defined by T (u[ ) = P(£u u[ ). Thus Euler’s equations can be written as ∂u[ /∂t + T (u[ ) = 0, which is in the “standard form” for an evolution equation. Note that T is nonlinear. Another important feature of T is that it is nonlocal; this is because P(α)(x) depends on the values of a on all of M and not merely those in the neighborhood of x ∈ M .  B.

We postulate an internal energy over the region W to be of the form Z Einternal = ρw dµ, W

9.2 Fluid Mechanics

555

where the function w is the internal energy density per unit mass. We assume that energy is balanced in the sense that the rate of change of energy in a region equals the work done on it: ! Z Z Z 1 d 2 kuk dµ + ρw dµ = − pu · n da. dt ϕt (W ) 2 ϕt (W ) ∂ϕt (W ) By the transport theorem and arguing as in our previous results, this reduces to   Z Dw dµ. 0= p div u + ρ dt ϕt (W ) Since W is arbitrary, Dw = 0. dt Now assume that w depends on the fluid motion through the density; that is, the internal energy depends only on how much the fluid is compressed. Such a fluid is called ideal isentropic or barotropic. The preceding identity then becomes   ∂w ∂ρ ∂w ∂w + dw · u = p div u + ρ +ρ dρ · u 0 = p div u + ρ ∂t ∂ρ ∂t ∂ρ ∂w = p div u + ρ (−ρ div u) ∂ρ p div u + ρ

using the equation of continuity. Since this is an identity and we are not restricting div u, we get p = ρ2 w0 (ρ). R If p is a given function of ρ note that w = − p d(1/ρ). In addition, dp/ρ = d(w + p/ρ). This follows from p = ρ2 w0 by a straightforward calculation in which p and w are regarded as functions of ρ. The quantity w + p/ρ = w + ρw0 is called the enthalpy and is often denoted h. Thus Euler’s equations for compressible ideal isentropic flow are ∂u[ 1 dp + £u u[ − d(u[ (u)) + = 0, ∂t 2 ρ ∂ρ + div(ρu) = 0, ∂t u(x, 0) = u0 (x) on M and u · n = 0 on ∂M. where p = ρ2 w0 (ρ) is a function of ρ, called an equation of state, which depends on the particular fluid. It is known that these equations lead to a well-posed initial value problem (i.e., there is a local existence and uniqueness theorem) only if p0 (ρ) > 0. This agrees with the common experience that increasing the surrounding pressure on a volume of fluid causes a decrease in occupied volume and hence an increase in density. Many gases can often be viewed as satisfying our hypotheses, with p = Aργ where A and γ are constants and γ ≥ 1.  Cases A and B above are rather opposite. For instance, if ρ = ρ0 is a constant for an incompressible fluid, then clearly p cannot be an invertible function of ρ. However, the case ρ = constant may be regarded as a limiting case p0 (ρ) → ∞. In Case B, p is an explicit function of ρ. In Case A, p is implicitly determined by the condition div u = 0. Finally, notice that in neither Case A or B is the possibility of a loss of total energy due to friction taken into account. This leads to the subject of viscous fluids, not dealt with here. Given a fluid flow with velocity field u(x, t), a streamline at a fixed time t is an integral curve of u; that is, if x(s) is a streamline parameterized by s at the instant t, then x(s) satisfies dx = u(x(s), t), ds

t fixed.

556

9. Applications

On the other hand, a trajectory is the curve traced out by a particle as time progresses, as explained at the beginning of this section; that is, is a solution of the differential equation dx = u(x(t), t) dt with given initial conditions. If u is independent of t (i.e., ∂u/∂t = 0), then, streamlines and trajectories coincide. In this case, the flow is called stationary or steady . This condition means that the “shape” of the fluid flow is not changing. Even if each particle is moving under the flow, the global configuration of the fluid does not change. The following criteria for steady solutions for homogeneous incompressible flow is a direct consequence of Euler’s equations, written in the form ∂u[ /∂t + P(£u u[ ) = 0, where P is the Hodge projection to the co-closed 1-forms. 9.2.2 Proposition. Let ut be a solution to the Euler equations for homogeneous incompressible flow on a compact manifold M and ϕt its flow. The following are equivalent: (i) ut is a steady flow (i.e., (∂u/∂t) = 0). (ii) ϕt is a one-parameter group: ϕt+s = ϕt ◦ ϕs . (iii) £u0 u[0 is an exact 1-form. (iv) iu0 du[0 is and exact 1-form. It follows from (iv) that if u0 is a harmonic vector field; that is, u0 satisfies δu[0 = 0 and du[0 = 0, then it yields a stationary flow. Also, it is known that there are other steady flows. For example, on a closed two-disk, with polar coordinates (r, θ), u = f (r)(∂/∂θ) is the velocity field of a steady flow because Z r (u · ∇)u = −∇p, where p(r, θ) = f 2 (s)s ds. 0

Clearly such a u need not be harmonic. We saw that for compressible ideal isentropic flow, the total energy  Z  kuk2 + ρw dµ 2 M is conserved. We can refine this a little for stationary flows as follows. 9.2.3 Theorem (Bernoulli’s Theorem). For stationary compressible ideal isentropic flow, with p a function of ρ, Z 1 1 p dp kuk2 + = kuk2 + w + 2 ρ 2 ρ R is constant along streamlines where the enthalpy dp/ρ = w + p/ρ denotes a potential for the one form dp/ρ. The same holds for stationary homogeneous (ρ = constant in space = ρ0 ) incompressible flow with R dp/ρ replaced by p/ρ0 . If body forces deriving from a potential U are present, that is, b[ = −dU, then the conserved quantity is Z 1 dp 1 p kuk2 + = kuk2 + w + + U. 2 ρ 2 ρ Proof. Since £u (u[ ) · u = d(u[ (u)) · u, for stationary ideal compressible or incompressible homogeneous flows we have 0=

∂u[ 1 dp · u = −(£u u[ ) · u + d(u[ ) · u − ·u ∂t 2 ρ 1 1 = − (dkuk2 ) · u − dp · u, 2 ρ

9.2 Fluid Mechanics

557

so that 

1 kuk2 + 2

Z

 x(s ) Z s2   Z 1 dp 2 dp 2 = d kuk + · u · x0 (s)ds ρ x(s1 ) 2 ρ s1 Z s2 ∂u[ · u(x(s))ds = 0 = ∂s s1

since x0 (s) = u(x(s)).



The two-form ω = du[ is called vorticity , which, in R3 can be identified with curl u. Our assumptions so far have precluded any tangential forces and thus any mechanism for starting or stopping rotation. Hence, intuitively, we might expect rotation to be conserved. Since rotation is intimately related to the vorticity, we can expect the vorticity to be involved. We shall now prove that this is so. Let C be a simple closed contour in the fluid at t = 0 and let Ct be the contour carried along the flow. In other words, Ct = ϕt (C) where ϕt is the fluid flow map. (See Figure 9.2.3.) The circulation around Ct is defined to be the integral Z u[ ,

ΓC t = C1

M

C

Figure 9.2.3. A loop advected by the flow

9.2.4 Theorem (Kelvin Circulation Theorem). Let M be a manifold and l ⊂ M a smooth closed loop, that is, a compact one-manifold. Let ut solve the Euler equations on M for ideal isentropic compressible or homogeneous incompressible flow and l(t) be the image of l at time t when each particle moves under the flow ϕt of ut ; that is, l(t) = ϕt (l). Then the circulation is constant in time; that is, Z d u[ = 0 dt l(t) t Proof.

Let ϕt be the flow of ut . Then l(t) = ϕt (l), and so changing variables, d dt

Z ϕt (l)

u[t =

 [  Z  ∂u ϕ∗t (£u u[ ) + ϕ∗t . ∂t l

However, £u u[ + ∂u[ /∂t is exact from the equations of motion and the integral of an exact form over a closed loop is zero. 

558

9. Applications

We now use Stokes’ theorem, which will bring in the vorticity. If Σ is a surface (a two-dimensional submanifold of M ) whose boundary is a closed contour C, then Stokes’ theorem yields Z Z Z [ [ ΓC = u = du = ω. C

Σ

Σ

See Figure 9.2.4.

C Figure 9.2.4. A surface and contour for Helmholtz’ theorem

Thus, as a corollary of the circulation theorem, we can conclude: 9.2.5 Theorem (Helmholtz’ Theorem). Under the hypotheses of Theorem 9.2.4, the flux of vorticity across a surface moving with the fluid is constant in time. We shall now show that ω and η = ω/ρ are Lie propagated by the flow. 9.2.6 Proposition. (i)

(ii)

For isentropic or homogeneous incompressible flow, we have ∂ω ∂η + £u ω = 0 and + £u η − η div u = 0 ∂t ∂t

called the vorticity–stream equation and ϕ∗t ω = ω0 and

ϕ∗t ηt = J(ϕt )η0

where ηt (x) = η(x, t) and J(ϕt ) is the Jacobian of ϕt . Proof.

Applying d to Euler’s equations for the two types of fluids we get the vorticity equation: ∂ω + £u ω = 0. ∂t

Thus     ∂η 1 ∂ω ω ∂ρ + £u η = + £u ω − 2 + dρ · u ∂t ρ ∂t p ∂t   η ∂ρ = + dρ · u + ρ div u + η div u = η div u ρ ∂t by conservation of mass. From ∂ω/∂t + £u ω = 0 it follows that (∂/∂t)(ϕ∗t ωt ) = 0, so ϕ∗t ωt = ω0 . Since ϕ∗t ρt = ρ0 /J(ϕt ) we also get ϕ∗t ηt = J(ϕt )η0 .  In three dimensions we can associate to η the vector field ζ = ∗η (or equivalently iζ µ = η). Thus ζ = curl u/ρ, if M is embedded in R3 . 9.2.7 Corollary.

If dim M = 3, then ζ is transported as a vector by ϕt ; that is, ζt = ϕt∗ ζ0

or

ζt (ϕt (x)) = Tx ϕt (ζt (x)).

9.2 Fluid Mechanics

Proof.

559

ϕ∗t ηt = J(ϕt )η0 by Proposition 9.2.6, so ϕ∗t iζt µ = J(ϕt )iζ0 µ.

But ϕ∗t iζt µ = iϕ∗t ζt ϕ∗t µ = iϕ∗t ζt J(ϕt )µ Thus iϕ∗t ζt u = iζ0 µ, which gives ϕ∗t ζt = ζ0 .



Notice that the vorticity as a two-form is Lie transported by the flow but as a vector field it is vorticity/ρ, which is Lie transported. Here is another instance where distinguishing between forms and vector fields makes an important difference. The flow ϕt of a fluid plays the role of a configuration variable and the velocity field u plays the role of the corresponding velocity variable. In fact, to understand fluid mechanics as a Hamiltonian system in the sense of §8.1, a first step is to set up its phase space using the set of all diffeomorphisms ϕ : M → M (volume preserving for incompressible flow) as the configuration space. The references noted at the beginning of this section carry out this program (see also Exercise 9.2-9 and §8.4).

Exercises  9.2-1.

In classical texts on fluid mechanics, the identity (u · ∇)u =

1 ∇(u · u) + (∇ × u) × u 2

is often used. To what identity does this correspond in this section?  9.2-2. A flow is called potential flow if u[ = dϕ for a function ϕ. For (not necessarily stationary) homogeneous incompressible or isentropic flow prove Bernoulli’s law in the form Z ∂ϕ 1 dp + kuk2 + = constant on a streamline. ∂t 2 ρ  9.2-3. Complex variables texts “show” that the gradient of ϕ(r, θ) = (r + 1/r) cos θ describes stationary ideal incompressible flow around a cylinder in the plane. Verify this in the context of this section.  9.2-4.

Translate Proposition 9.2.2 into vector analysis notation in R3 and give a direct proof.

 9.2-5.

Let dim M = 3, and assume the vorticity ω has a one-dimensional kernel.

(i) Using Frobenius’ theorem, show that this distribution is integrable. (ii) Identify the one-dimensional leaves with integral curves of ζ (see Corollary 9.2.7)—these are called vortex lines. (iii) Show that vortex lines are propagated by the flow.  9.2-6. Assume dim M = 3. A vortex tube T is a closed oriented two-manifold in M that is a union of vortex lines. The strength of the vortex tube is the flux of vorticity across a surface Σ inside T whose boundary lies on T and is transverse to the vortex lines. Show that vortex tubes are propagated by the flow and have a strength that is constant in time.  9.2-7. Let f : R3 → R be a linear function and g : S 2 → R be its restriction to the unit sphere. Show that dg gives a stationary solution of Euler’s equations for flow on the two-sphere.  9.2-8.

Stream Functions

560

9. Applications

(i) For incompressible flow in R2 , show that there is a function ψ such that u1 = ∂ψ/∂y and u2 = −∂ψ/∂x. One calls ϕ the stream function (as in Batchelor [1967]). (ii) Show that if we let ∗ ψ = ψ dx ∧ dy be the associated two form, then u[ = δ ∗ ψ. (iii) Show that u is a Hamiltonian vector field (see §8.1) with energy ψ directly in R2 and then for arbitrary two-dimensional Riemannian manifolds M . (iv) do stream functions exist for arbitrary fluid flow on T2 ? On S 2 ? (v) Show that the vorticity is ω = ∆∗ ψ.  9.2-9 (Clebsch Variables; Clebsch [1859]). Let F be the space of functions on a compact manifoldR M with the dual space F ∗ , taken to be densities on M ; the pairing between f ∈ F and ρ ∈ F ∗ is hf, ρi = M f ρ. (i) On the symplectic manifold F ×F ∗ ×F ×F ∗ with variables (α, λ, µ, ρ), show that Hamilton’s equations for a given Hamiltonian H are α˙ =

δH , δλ

µ˙ =

δH , δρ

δH λ˙ = − , δα

ρ˙ = −

δH , δµ

where δH/δλ is the functional derivative of H defined by   δH ˙ ˙ , λ = DH(λ) · λ. δλ (ii) In the ideal isentropic compressible fluid equations, set M = ρu[ , the momentum density , where dx denotes the Riemannian volume form on M. Identify the density σ(x)dx ∈ F ∗ with the function σ(x) ∈ F and write M = −(ρdµ+λdα)dx. For momentum densities of this form show that Hamilton’s equations written in the variables (α, λ, µ, ρ) imply Euler’s equation and the equation of continuity.

9.3

Electromagnetism

Classical electromagnetism is governed by Maxwell’s field equations. The form of these equations depends on the physical units chosen, and changing these units introduces factors like 4π, c = the speed of light, 0 = the dielectric constant and µ0 = the magnetic permeability. The discussion in this section assumes that 0 , µ0 are constant; the choice of units is such that the equations take the simplest form; thus c = 0 = µ0 = 1 and factors 4π disappear. We also do not consider Maxwell’s equations in a material, where one has to distinguish E from D, and B from H. Let E, B, and J be time dependent C 1 -vector fields on R3 and ρ : R3 × R → R a scalar. These are said to satisfy Maxwell’s equations with charge density ρ and current density J when the following hold: div E = ρ (Gauss’s law ) div B = 0 (no magnetic sources) ∂B curl E + = 0 (Faraday’s law of induction) ∂t ∂E curl B − = J (Amp` ere’s law ) ∂t

(9.3.1) (9.3.2) (9.3.3) (9.3.4)

E is called the electric field and B the magnetic field . R The quantity Ω ρ dV = Q is called the charge of the set Ω ⊂ R3 . By Gauss’ theorem, equation (9.3.1) is equivalent to Z Z E · n dS = ρ dV = Q (9.3.5) ∂Ω

Ω

9.3 Electromagnetism

561

for any (nice) open set Ω ⊂ R3 ; that is, the electric flux out of a closed surface equals the total charge inside the surface. This generalizes Gauss’ law for a point charge discussed in §7.3. By the same reasoning, equation (9.3.2) is equivalent to Z B · n dS = 0.

(9.3.6)

∂Ω

That is, the magnetic flux out of any closed surface is zero. In other words there are no magnetic sources inside any closed surface. By Stokes’ theorem, equation (9.3.3) is equivalent to Z Z Z ∂ B · n dS (9.3.7) E · ds = (curl E) · n dS = − ∂t S ∂S S R for any closed loop ∂S bounding a surface S. The quantity ∂S E · ds is called the voltage around ∂S. Thus, Faraday’s law of induction equation (9.3.3), says that the voltage around a loop equals the negative of the rate of change of the magnetic flux through the loop. Finally, again by the classical Stokes’ theorem, equation (9.3.4) is equivalent to Z Z Z Z ∂ E · n dS + J · n dS. (9.3.8) B · ds = (curl B) · n dS = ∂t S S ∂S S R Since S J · n dS has the physical interpretation of current, Amp`ere’s law states that if E is constant in R time, then the magnetic potential difference ∂S B · ds around a loop equals the current through the loop. In general, if E varies in time, Amp`ere’s law states that the magnetic potential difference around a loop equals the total current in the loop plus the rate of change of electric flux through the loop. We now show how to express Maxwell’s equations in terms of differential forms. Let M = R4 = {(x, y, z, t)} with the Lorentz metric g on R4 having diagonal form (1, 1, 1, −1) in standard coordinates (x, y, z, t). 9.3.1 Proposition.

There is a unique two-form F on R4 , called the Faraday two-form such that E[ = −i∂/∂t F ; B[ = −i∂/∂t ∗ F.

(Here the in R4 .) Proof.

[

(9.3.9) (9.3.10)

is associated with the Euclidean metric in R3 and the ∗ is associated with the Lorentzian metric

If F = Fxy dx ∧ dy + Fzx dz ∧ dx + Fyz dy ∧ dz + Fxt dx ∧ dt + Fyt dy ∧ dt + Fzt dz ∧ dt,

then (see Example 7.2.14E), ∗F = Fxy dz ∧ dt + Fzx dy ∧ dt + Fyz dx ∧ dt − Fxt dy ∧ dz − Fyt dz ∧ dx − Fzt dx ∧ dy and so −i∂/∂t F = Fxt dx + Fyt dy + Fzt dz and −i∂/∂t ∗ F = Fxy dz + Fzx dy + Fyz dx. Thus, F is uniquely determined by equations (9.3.9) and (9.3.10), namely F = E 1 dx ∧ dt + E 2 dy ∧ dt + E 3 dz ∧ dt + B 3 dx ∧ dy + B 2 dz ∧ dx + B 1 dy ∧ dz.



562

9. Applications

We started with E and B and used them to construct F , but one can also take F as the primitive object and construct E and B from it using equations (9.3.9) and (9.3.10). Both points of view are useful. Similarly, out of ρ and J we can form the source one-form j = −ρ dt + J1 dx + J2 dy + J3 dz; that is, j is uniquely determined by the equations −i∂/∂t j = ρ and i∂/∂t ∗ j = ∗J[ ; in the last relation, J is regarded as being defined on R4 . 9.3.2 Proposition.

Maxwell’s equations (9.3.1)–(9.3.4) are equivalent to the equations dF = 0

and

δF = j

on the manifold R4 endowed with the Lorentz metric. Proof.

A straightforward computation shows that     ∂B ∂B dF = curl E + dy ∧ dz ∧ dt + curl E + dz ∧ dx ∧ dt ∂t x ∂t y   ∂B dx ∧ dy ∧ dt + (div B)dx ∧ dy ∧ dz. + curl E + ∂t z

Thus dF = 0 is equivalent to equations (9.3.2) and (9.3.3). Since the index of the Lorentz metric is 1, we have δ = ∗d∗. Thus, δF = ∗d ∗ F = ∗d −E 1 dy ∧ dz − E 2 dz ∧ dx − E 3 dx ∧ dy
+ B 1 dx ∧ dt + B 2 dy ∧ dt + B 3 dz ∧ dt    ∂B dy ∧ dz ∧ dt + = ∗ −(div E)dx ∧ dy ∧ dz + curl B − ∂t x #     ∂E ∂E + curl B − dz ∧ dx ∧ dt + curl B − dx ∧ dy ∧ dt ∂t y ∂t z     ∂E ∂E = curl B − dx + curl B − dy ∂t x ∂t y   ∂E + curl B − dz − (div E) dt. ∂t z Thus δF = j iff equations (9.3.1) and (9.3.4) hold. As a skew matrix, we can represent F as follows  0 B3 −B 3 0 F =  B 2 −B 1 −E 1 −E 2



−B 2 B1 0 −E 3

 E1 E2 . E3 0

Recall from §7.5 and Exercise 8.5-7, the formula (δF )i = | det[glj ]|−1/2 (F ik | det[glj ]|1/2 ),k Since | det[gkl ]| = 1, Maxwell’s equations can be written in terms of the Faraday two-form F in components as Fij,k + Fjk,i + Fki,j = 0 (9.3.11) and ik F,k = −j i ,

(9.3.12)

9.3 Electromagnetism

563

where Fij,k = ∂Fij /∂xk , etc. Since δ 2 = 0, we obtain 0 = δ 2 F = δj = ∗d ∗ j = ∗d(−ρ dx ∧ dy ∧ dz + (∗J[ ) ∧ dt)    ∂ρ ∂ρ + div J dx ∧ dy ∧ dz ∧ dt = + div J; =∗ ∂t ∂t that is, ∂ρ/∂t + div J = 0, which is the continuity equation (see §8.2). Its integral form is, by Gauss’ theorem, Z Z dQ d = ρ dV = J · n dS dt dt Ω ∂Ω for any bounded open set Ω. Thus the continuity equation says that the flux of the current density out of a closed surface equals the rate of change of the total charge inside the surface. Next we show that Maxwell’s equations are Lorentz invariant, that is, are special-relativistic. The Lorentz group £ is by definition the orthogonal group with respect to the Lorentz metric g, that is,  £ = A ∈ GL(R4 ) | g(Ax, Ay) = g(x, y) for all x, y ∈ R4 . Lorentz invariance means that F satisfies Maxwell’s equations with j iff A∗ F satisfies them with A∗ F , for any A ∈ £. But due to Proposition 9.3.1 this is clear since pull-back commutes with d and orthogonal transformations commute with the Hodge operator (see Exercise 7.2-4) and thus they commute with δ. As a 4 × 4 matrix, the Lorentz transformation A acts on F by F 7→ A∗ F = AF AT . Let us see that the action of A ∈ £ mixes up E’s and B’s. (This is the source of statements like: “A moving observer sees an electric field partly converted to a magnetic field.”) Proposition 9.3.1 defines E and B intrinsically in terms of F . Thus, if one performs a Lorentz transformation A on F , the new resulting electric and magnetic fields E0 and B0 with respect to the Lorentz unit normal A∗ (∂/∂t) to the image A(R3 × 0) in R4 are given by (E0 )[ = −iA∗ ∂/∂t A∗ F, For a Lorentz transformation of the form x − vt , x0 = √ 1 − v2

y 0 = y,

(B0 )[ = −iA∗ ∂/∂t A∗ F.

z 0 = z,

t − vx t0 = √ 1 − v2

(the special-relativistic analogue of an observer moving uniformly along the x-axis with velocity v) we get   2 3 E 3 + vB 2 0 0 E − vB E = E, √ , √ 1 − v2 1 − v2 and 0

B =



B 2 − vE 3 B 3 + vE 2 B, √ , √ 1 − v2 1 − v2 0

 .

We leave the verification to the reader. By the way we have set things up, note that Maxwell’s equations make sense on any Lorentz manifold; that is, a four-dimensional manifold with a pseudo-Riemannian metric of signature (+, +, +, −). Maxwell’s vacuum equations (i.e., j = 0) will now be shown to be conformally invariant on any Lorentz manifold (M, g). A diffeomorphism ϕ : (M, g) → (M, g) is said to be conformal if ϕ∗ g = f 2 g for a nowhere vanishing function f . (See Fulton, Rohrlich, and Witten [1962] for a review of conformal invariance in physics and the original literature references.) 9.3.3 Proposition. Let F ∈ Ω2 (M ) where (M, g) is a Lorentz manifold, satisfy dF = 0 and δF = j. Let ϕ be a conformal diffeomorphism. Then ϕ∗ F satisfies dϕ∗ F = 0

and

δϕ∗ F = f 2 ϕ∗ j.

Hence Maxwell’s vacuum equations (with j = 0) are conformally invariant; that is, if F satisfies them, so does ϕ∗ F .

564

9. Applications

Proof. Since ϕ∗ commutes with d, dF = 0 implies dϕ∗ F = 0. The second equation implies ϕ∗ δF = ϕ∗ j. By Exercise 8.5-8, we have δϕ∗ g ϕ∗ β = ϕ∗ δβ. Hence δF = j implies δϕ∗ g ϕ∗ F = ϕ∗ j = δf 2 g ϕ∗ F since ϕ is conformal. The local formula for δF, namely (δF )i = | det[gks ]|−1/2 gir

∂ (g ra g lb Fab | det[gks ]|−1/2 ) ∂xl

shows that when one replaces g by f 2 g, we get δf 2 g ϕ∗ F = f −2 ϕ∗ F, and so δϕ∗ F = f 2 ϕ∗ j.



Let us now discuss the energy equation for the electromagnetic field. Introduce the energy density of the field 1 2 E = (E · E + B · B) and the Poynting energy-flux vector S = E × B. Poynting’s theorem states that −

∂E = div S + E · J. ∂t

This is a straightforward computation using equations (9.3.3) and (9.3.4). We shall extend this result to R4 and, at the same time, shall rephrase it in the framework of forms. Introduce the stress–energy–momentum tensor (or the Maxwell stress tensor ) T by T ij = F ik Fkj − 41 g ij Fpq F pq

(9.3.13)

(or intrinsically, T = F · F − 41 hF, F ig, where F · F denotes a single contraction of F with itself). A straightforward computation shows that the divergence of T equals j ,i pq T,jij = F,jik Fkj + F ik Fk,j − 12 Fpq F ,i where Fpq = (∂Fpq /∂xk )g ik . Taking into account δF = j written in the form (9.3.12), it follows that

T,lil = F ik jk .

(9.3.14)

For i = 4, the relation (9.3.14) becomes Poynting’s theorem.2 It is clear that T is a symmetric 2-tensor. As a symmetric matrix,   σ E×B T = , (E × B)T E 2 Poynting’s

theorem can also be understood in terms of a Hamiltonian formulation; see Example 9.4.2 below. The Poynting energy-flux vector is the Noether conserved quantity for the action of the diffeomorphism group of R3 on T ∗ A, where A is the space of vector potentials A defined in the following paragraph, and Poynting’s theorem is just conservation of momentum (Noether’s theorem). We shall not dwell upon these aspects and refer the interested reader to Abraham and Marsden [1978] and Marsden and Ratiu [1999].

9.3 Electromagnetism

565

where σ is the stress tensor and E is the energy density. The symmetric 3 × 3 matrix σ has the following components σ 11 = 21 [−(E 1 )2 − (B 1 )2 + (E 2 )2 + (B 2 )3 + (E 3 )2 + (B 3 )2 ] σ 22 = 21 [(E 1 )2 + (B 1 )2 − (E 2 )2 − (B 2 )2 + (E 3 )3 + (B 3 )2 ] σ 33 = 21 [(E 1 )2 + (B 1 )2 + (E 2 )2 + (B 2 )2 − (E 3 )2 − (B 3 )2 ] σ 12 = E 1 E 2 + B 1 B 2 σ 13 = E 1 E 3 + B 1 B 3 σ 23 = E 2 E 3 + B 2 B 3 . We close this section with a discussion of Maxwell’s equations in terms of vector potentials. We first do this directly in terms of E and B. Since div B = 0, if B is smooth on all of R3 , there exists a vector field A, called the vector potential , such that B = curl A, by the Poincar´e lemma. This vector field A is not unique and one could also use A0 = A + grad f for some (possibly time–dependent) function f : R3 → R. This freedom in the choice of A is called gauge freedom. For any such choice of A we have by equation (9.3.3)   ∂ ∂A ∂B = curl E + curl A = curl E + , 0 = curl E + ∂t ∂t ∂t so that again by the Poincar´e lemma there exists a (time-dependent) function ϕ : R3 → R such that E+

∂A = − grad ϕ. ∂t

(9.3.15)

Recall that the Laplace–Beltrami operator on functions is defined by ∇2 f = div(grad f ). On vector fields in R3 this operator may be defined componentwise. Then it is easy to check that curl(curl A) = grad(div A) − ∇2 A. Using this identity, (9.3.15), and B = curl A in (9.3.4), we get   ∂E ∂ ∂A J = curl B − = curl(curl A) − − − grad ϕ ∂t ∂t ∂t ∂2A ∂ = grad(div A) − ∇2 A + + (grad ϕ), ∂t2 ∂t and thus

  ∂2A ∂ϕ ∇ A− = −J + grad div A + . ∂t2 ∂t 2

(9.3.16)

From equation (9.3.1) we obtain as before   ∂A ∂ ρ = div E = div − − grad ϕ = −∇2 ϕ − (div A), ∂t ∂t that is, ∇2 ϕ = −ρ −

∂ (div A), ∂t

or subtracting ∂ 2 ϕ/∂t2 from both sides, ∇2 ϕ −

∂2ϕ ∂ = −ρ − + 2 ∂t ∂t

 div A +

∂ϕ ∂t

 .

(9.3.17)

566

9. Applications

It is apparent that equations (9.3.16) and (9.3.17) can be considerably simplified if one could choose, using the gauge freedom, the vector potential A and the function ϕ such that div A +

∂ϕ = 0. ∂t

Assume one has chosen A0 , ϕ0 and one seeks a function f such that A = A0 + grad f and ϕ = ϕ0 − ∂f /∂t satisfy div A + ∂ϕ/∂t = 0. This becomes, in terms of f ,   ∂ ∂f ∂ϕ0 ∂2f 0 = div(A0 + grad f ) + ϕ0 − = div A0 + + ∇2 f − 2 ; ∂t ∂t ∂t ∂t that is,   ∂ϕ0 ∂2f ∇ f − 2 = − div A0 + . ∂t ∂t 2

(9.3.18)

This equation is the classical inhomogeneous wave equation. The homogeneous wave equation (righthand side equals zero) has solutions f (t, x, y, z) = ψ(x − t) for any function ψ. This solution propagates the graph of ψ like a wave—hence the name wave equation. Now we we can draw some conclusions regarding Maxwell’s equations. In terms of the vector potential A and the function ϕ, equations (9.3.1) and (9.3.4) become ∇2 ϕ −

∂2ϕ = −ρ, ∂t2

∇2 A −

∂2A = −J, ∂t2

(9.3.19)

which again are inhomogeneous wave equations. Conversely, if A and ϕ satisfy the foregoing equations and div A + ∂ϕ/∂t = 0, then E = − grad ϕ − ∂A/∂t and B = curl A satisfy Maxwell’s equations. Thus in R4 , this procedure reduces the study of Maxwell’s equations to the wave equation, and hence solutions of Maxwell’s equations can be expected to be wavelike. We now repeat the foregoing constructions on R4 using differential forms. Since dF = 0, on R4 we can write F = dG for a one-form G. Note that F is unchanged if we replace G by G + df. This again is the gauge freedom. Substituting F = dG into δF = j gives δdG = j. Since ∆ = dδ + δd is the Laplace–deRham operator in R4 , we get ∆G = j − dδG. (9.3.20) Suppose we try to choose G so that δG = 0 (a gauge condition). To do this, given an initial G0 , we can let G = G0 + df and demand that 0 = δG = δG0 + δdf = δG0 + ∆f so f must satisfy ∆f = −δG0 . Thus, if the gauge condition ∆f = −δG0

(9.3.21)

∆G = j.

(9.3.22)

holds, then Maxwell’s equations become Equation (9.3.21) is equivalent to (9.3.18) and (9.3.22) to (9.3.19) by choosing G = A[ + ϕ dt (where Euclidean in R3 ).

Exercises  9.3-1.

Assume that the Faraday two-form F depends only on t − x.

(i) Show that dF = 0 is then equivalent to B 3 = E 2 , B 2 = −E 3 , B 1 = 0.

[

is

9.3 Electromagnetism

567

(ii) Show that δF = 0 is then equivalent to B 3 = E 2 , B 2 = −E 3 , E 1 = 0. These solutions of Maxwell’s equations are called plane electromagnetic waves; they are determined only by E 2 , E 3 or B 2 , B 3 , respectively.  9.3-2. Let u = ∂/∂t. Show that the Faraday two-form F ∈ Ω2 (R4 ) is given in terms of E and B ∈ X(R4 ) by F = u[ ∧ E [ − ∗(u[ ∧ B [ ).  9.3-3.

Show that the Poynting vector satisfies S [ = ∗(B [ ∧ E [ ∧ u[ )

where u = ∂/∂t and E, B ∈ X(R4 ).  9.3-4. Let (M, g) be a Lorentzian four-manifold and u ∈ X(M ) a timelike unit vector field on M ; that is, g(u, u) = −1. (i) Show that any α ∈ Ω2 (M ) can be written in the form α = (iu α) ∧ u[ − ∗((iu ∗ α) ∧ u[ ). (ii) Show that if iu α = 0, where α ∈ Ω2 (M ) (“α is orthogonal to u”), then ∗α is decomposable, that is, ∗α is the wedge product of two one-forms. Prove that α is also locally decomposable. Hint: Use the Darboux theorem.  9.3-5.

The field of a stationary point charge is given by er , B = 0, E= 4πr3

where r is the vector xi + yj + zk in R3 and r is its length. Use this and a Lorentz transformation to show that the electromagnetic field produced by a charge e moving along the x-axis with velocity v is E=

e (1 − v 2 )r 2 4π [x + (1 − v 2 )(y 2 + z 2 )]exc:3.2−27

and, using spherical coordinates with the x-axis as the polar axis, Br = 0,

Bθ = 0,

Bϕ =

e(1 − v 2 )v sin θ 4πr2 (1 − v 2 sin2 θ)exc:3.2−27

(the magnetic field lines are thus circles centered on the polar axis and lying in planes perpendicular to it).  9.3-6 ( Misner, Thorne, and Wheeler [1973]). The following is the Faraday two-form for the field of an electric dipole of magnitude p1 oscillating up and down parallel to the z-axis.     1 iω F = Re p1 eiωr−iωt 2 cos θ − dr ∧ dt r2 r2   1 iω ω 2 + sin θ − − rdθ ∧ dt r3 r2 r    iω ω 2 + sin θ − 2 − dr ∧ r dθ . r r Verify that dF = 0 and δF = 0, except at the origin.  9.3-7.

Let the Lagrangian for electromagnetic theory be p 1 £ = |F |2 = − Fij Fkl g ik g jl − det g. 2

Check that ∂£/∂gij is the stress–energy–momentum tensor T ij (see Hawking and Ellis [1973, Section 3.3]).

568

9.4

9. Applications

The Lie–Poisson Bracket in Continuum Mechanics and Plasma Physics

This section studies the equations of motion for some Hamiltonian systems in Poisson bracket formation. As opposed to §8.1, the emphasis is placed here on the Poisson bracket rather than on the underlying symplectic structure. This naturally leads to a generalization of Hamiltonian mechanics to systems whose phase space is a “Poisson manifold.” We do not intend to develop here the theory of Poisson manifolds but only to illustrate it with the most important example, the Lie–Poisson bracket. See Marsden and Ratiu [1999] for further details. If (P, ω) is a (weak) symplectic manifold, H : P → R a smooth Hamiltonian with Hamiltonian vector field XH ∈ X(P ) whose flow is denoted by ϕt , recall from Corollary 9.1.11 that dϕt (p) = XH (ϕt (p)) dt

(9.4.1)

is equivalent to d (F ◦ ϕt ) = {F ◦ ϕt , H ◦ ϕt } (9.4.2) dt for any smooth locally defined function f : U → R, where U is open in P . In (9.4.2), { , } denotes the Poisson bracket defined by ω, that is, {F, G} = ω(XF , XG ) = XG [F ] = −XF [G].

(9.4.3)

Finally, recall that the Poisson bracket is an antisymmetric bilinear operation on F(P ) which satisfies the Jacobi identity {{F, G}, H} + {{G, H}, F } + {{H, F }, G} = 0, (9.4.4) that is, (F(P ), { , }) is a Lie algebra. In addition, the multiplicative ring structure and the Lie algebra structure of F(P ) are connected by the Leibniz rule {F G, H} = F {G, H} + G{F, H},

(9.4.5)

that is, { , } is a derivation in each argument. These observations naturally lead to the following generalization of the concept of symplectic manifolds. 9.4.1 Definition. A smooth manifold P is called a Poisson manifold if F(P ), the ring of functions on P , admits a Lie algebra structure which is a derivation in each argument. The bracket operation on F(P ) is called a Poisson bracket and is usually denoted by { , }. From the remarks above, we see that any (weak) symplectic manifold is a Poisson manifold. One of the purposes of this section is to show that there are physically important Poisson manifolds which are not symplectic. But even in the symplectic context it is sometimes easier to compute the Poisson bracket than the symplectic form, as the following example shows. 9.4.2 Example (Maxwell’s Vacuum Equations as a Hamiltonian System). We shall indicate how the dynamical pair of Maxwell’s vacuum equations (9.3.3) and (9.3.4) of the previous section with current J = 0 are a Hamiltonian system. As the configuration space for Maxwell’s equations, we take the space A of vector potentials. (In more general situations on spaces other than R3 , one should replace A by the set of connections on a principal bundle over configuration space.) The corresponding phase space is then the cotangent bundle T ∗ A with the canonical symplectic structure. Elements of T ∗ A may be identified with pairs (A, Y) where Y is a vector field density on R3 . (We do not distinguish Y and Y d3 x.) The pairing between A’s and Y’s is given by integration, so the canonical symplectic structure ω on T ∗ A is Z ω((A1 , Y1 ), (A2 , Y2 )) = (Y2 · A1 − Y1 · A2 )d3 x, (9.4.6) R3

9.4 The Lie–Poisson Bracket in Continuum Mechanics and Plasmas

569

with associated Poisson bracket Z {F, G}(A, Y) = R3



δF δG δF ∂G · − · δA δY δY δA



d3 x,

(9.4.7)

where δF/δA is the vector field defined by DA F (A, Y) · A0 =

Z

δF · A0 d3 x. δA

with the vector field δF/δY defined similarly. With the Hamiltonian Z 1 1 kYk2 d3 x + k curl Ak2 d3 x, H(A, Y) = 2 2

(9.4.8)

Hamilton’s equations are easily computed to be ∂Y = − curl curl A ∂t

and

∂A = Y. ∂t

If we write B for curl A and E for −Y, the Hamiltonian becomes the field energy Z 1 1 kEk2 d3 x + kBk3 d3 x. 2 2

(9.4.9)

(9.4.10)

Equation (9.4.9) implies Maxwell’s equations ∂E = curl B ∂t

and

∂B = − curl E, ∂t

and the Poisson bracket of two functions F (A, E), G(A, E) is  Z  δF δG δG δF {F, G}(A, E) = − · − · d3 x. δA δE δA δE R3

(9.4.11)

(9.4.12)

We can express this Poisson bracket in terms of E and B = curl A. To do this, we consider functions F˜ : V × X(R3 ) → R, where  V = curl Z | Z ∈ X(R3 ) . We pair V with itself relative to the L2 -inner product. This is a weakly non-degenerate pairing since by the Hodge–Helmholtz decomposition Z curl Z1 · curl Z2 d3 x = 0 R3 3

for all Z2 ∈ X(R ) implies that curl Z1 = ∇f, whence ∆f = div ∇f = div curl Z1 = 0, and so, f = constant by Liouville’s theorem. Therefore curl Z1 = ∇f = 0, as was to be shown. We compute δF/δA in terms of the functional derivative of an arbitrary extension Fˆ of F˜ to X(R3 ), where F (A, E) = F˜ (B, E), for B = curl A. Let L be the linear map L(A) = curl A so that F = F˜ ◦ (L × Identity) = Fˆ ◦ (L × Identity). By the chain rule, we have for any δA ∈ X(R3 ), Z δF · δAd3 x = DF (A) · δA = (DFˆ (B) ◦ DL(A)) · δA δA 3 R Z Z δ Fˆ δ Fˆ 3 = · curl δA d x = curl · δAd3 x, δB R3 δB R3

570

9. Applications

since DL(A) = L and

R R3

X · curl Yd3 x =

R R3

Y · curl X d3 x. Therefore δ F˜ δF = curl . δA δB

(9.4.13)

This formula seems to depend on the extension Fˆ of F˜ . However, this is not the case. More precisely, let K : X(R3 ) × X(R3 ) → R be such that K|(V × X(R)) ≡ 0. We claim that if B ∈ V, then δK/δB is a gradient. Granting this statement, this shows that equation (9.4.13) is independent of the extension, since any two extensions of F˜ coincide on V × X(R3 ) and since curl ◦ grad = 0. To prove the claim, note that for any Z ∈ X(R3 ), Z Z δK δK 3 · curl Z d x = · Z d3 x 0 = DK(B) · curl Z = curl δB δB R3 R3 whence curl δK/δB = 0, that is, δK/δB is a gradient. Thus, equation(9.4.13) implies δF δ F˜ = curl , δA δB

(9.4.14)

where on the right-hand side δ F˜ /δB is understood as the functional derivative relative to B of an arbitrary extension of F˜ to X(R3 ). Since δ F˜ /δE = δF/δE, the Poisson bracket (9.4.12) becomes ! Z ˜ ˜ ˜ ˜ δ G δ G δ F δ F ˜ · curl − · curl d3 x (9.4.15) {F˜ , G}(B, E) = δE δB δE δB 3 R This bracket was found by Born and Infeld [1935] by a different method. Using the Hamiltonian Z 1 H(B, E) = (kBk2 + kEk2 )d3 x, 2 R3

(9.4.16)

equations (9.4.11) are equivalent to the Poisson bracket equations ˜ F˜˙ = {F˜ , H}.

(9.4.17)

˜ ˜ Indeed, since δ H/δE = E, curl(δ H/δB) = curl B, we have ˜ = {F˜ , H}

Z R3

Z = R3

δ F˜ δ F˜ · curl B − E · curl δE δB

!

δ F˜ δ F˜ · curl B − E · curl δE δB

!

d3 x d3 x

Moreover, ˙ + DF˜ (E) · E ˙ F˜ = DF˜ (B) · B ! Z δ F˜ ˙ δ 0 F˜ ˙ = ·B+ · E d3 x δB δE R3 ! Z δ F˜ ˙ δ F˜ ˙ = ·B+ · E d3 x, δB δE 3 R where δ 0 F˜ /δB denotes the functional derivative of F˜ relative to B in V, that is, Z δ 0 F˜ DF˜ (B) · δB = · δB d3 x. R3 δB

(9.4.18)

9.4 The Lie–Poisson Bracket in Continuum Mechanics and Plasmas

571

The last equality in the formula for F˜˙ is proved in the following way. Recall that δ F˜ /δB is the functional derivative of an arbitrary extension of F˜ computed at B, that is, Z ˜ δF DF˜ (B) · Z = · Z d3 x for any Z ∈ X(R3 ); δB therefore since δB is a curl, this implies δ 0 F˜ /δB and δ F˜ /δB differ by a gradient which is L2 -orthogonal to ˙ since B ˙ is divergence free (again by the Helmholtz–Hodge decomposition). Therefore equation (9.4.11) B, holds if and only if equation (9.4.17) does.  The Lie–Poisson bracket. We next turn to the most important example of a Poisson manifold which is not symplectic. Let g denote a Lie algebra that is, a vector space with a pairing [ξ, η] of elements of g that is bilinear, antisymmetric and satisfies Jacobi’s identity. Let g∗ denote its “dual”, that is, a vector space weakly paired with g via h , i : g∗ × g → R. If g is finite dimensional, we take this pairing to be the usual action of forms on vectors. 9.4.3 Definition.

For F, G : g∗ → R, define the (±) Lie–Poisson brackets by    δF δG , {F, G}± (µ) = ± µ, δµ δµ

(9.4.19)

where µ ∈ g∗ and δF/δµ, δG/δµ ∈ g are the functional derivatives of F and G, that is, DF (µ) · δµ = hδµ, δF/δµi. If g is finite dimensional with a basis ξi , and the structure constants are defined by [ξi , ξj ] = ckij ξk , the Lie–Poisson bracket is {F, G} = ±µji cijk 9.4.4 Theorem (Lie–Poisson Theorem). manifold.

δF δG . δµj δµk

The dual space g∗ with the (±) Lie–Poisson bracket is a Poisson

Proof. Clearly { , }± is bilinear and skew symmetric. To show { , }± is a derivation in each argument, we show that δ(F G) δG δF = F (µ) + G(µ) . (9.4.20) δµ δµ δµ To prove (9.4.20), let δµ ∈ g∗ be arbitrary. Then   δ(F G) δµ, = D(F G)(µ) · δµ δµ = F (µ)DG(µ) · δµ + G(µ)DF (µ) · δµ   δG δF = δµ, F (µ) + G(µ) . δµ δµ Finally, we prove the Jacobi identity. We start by computing the derivative of the map µ ∈ g∗ 7→ δF/δµ ∈ g. We have for every λ, ν ∈ g∗   δF D ν, (µ) · λ = D(DF (·) · ν)(µ) · λ = D2 F (µ)(ν, λ), δµ that is,  D

δF δµ



(µ) · λ = D2 F (µ)(λ, ·).

(9.4.21)

572

9. Applications



 δF δG Therefore the derivative of µ 7→ , is δµ δµ       δF δG δG δF 2 2 D , (µ) · ν = D F (µ)(ν, ·), + , D G(µ)(ν, ·) δµ δµ δµ δµ

(9.4.22)

where D2 F (µ)(ν, ·) ∈ L(g∗ , R) is assumed to be represented via h , i by an element of g. Therefore by (9.4.19) and (9.4.22)   δ ν, {F, G} = D{F, G}(µ) · ν δµ       δF δG δG 2 = ν, , + µ, D F (µ)(ν, ·), δµ δµ δµ    δF + µ, , D2 G(µ)(ν, ·) δµ      ∗  δF δG δG = ν, , − ad µ, D2 F (µ)(ν, ·) δµ δµ δµ ∗    δF µ, D2 G(µ)(ν, ·) , + ad δµ where ad(ξ) : g → g is the linear map ad(ξ) · η = [ξ, η] and ad(ξ)∗ : g∗ → g∗ is its dual defined by had(ξ)∗ µ, ηi = hµ, [ξ, ν]i,

η ∈ g, µ ∈ g∗ .

Therefore,         ∗  δF δG δ δG 2 , ν, {F, G} = ν, − ν, D F (µ) ad µ, · δν δµ δµ δµ    ∗  δF + ν, D2 G(µ) ad µ, · δµ     ∗  δF δG δ δG {F, G} = , − D2 F (µ) ad µ, · δν δµ δµ δµ   ∗  δF µ, · , + D2 G(µ) ad δµ

(9.4.23)

which in turn implies        δH δF δG δH δ {F, G}, = µ, , , {{F, G}, (µ)H} = µ, δµ δµ δµ δµ δµ   ∗  ∗  δG δH + D2 F (µ) ad µ, ad µ δµ δµ   ∗  ∗  δF δH µ, ad µ , − D2 G(µ) ad δµ δµ The two cyclic permutations in F, G, H added to the above formula sum up to zero: all six terms involving second derivatives cancel and the three first terms add up to zero by the Jacobi identity for the bracket of g.  9.4.5 Example (The Free Rigid Body). The equations of motion of the free rigid body described by an observer fixed on the moving body are given by Euler’s equation ˙ = Π × ω, Π

(9.4.24)

9.4 The Lie–Poisson Bracket in Continuum Mechanics and Plasmas

573

where Π, ω ∈ R3 , Πi = Ii ωi , i = 1, 2, 3, I = (I1 , I2 , I3 ) are the principal moments of inertia, the coordinate system in the body is chosen so that the axes are the principal axes, ω is the angular velocity in the body, and Π is the angular momentum in the body. It is straightforward to check that the kinetic energy H(Π) =

1 Π·ω 2

(9.4.25)

is a conserved quantity for equation (9.4.24). We shall prove below that (9.4.24) are Hamilton’s equations with Hamiltonian (9.4.25) relative to a (−) Lie–Poisson structure on R3 . The vector space R3 is in fact a Lie algebra with respect to the bracket operation given by the cross product, that is, [x, y] = x × y. (This is the structure that it inherits from the rotation group.) We pair R3 with itself using the usual dot-product, that is, hx, yi = x · y. Therefore, if F : R3 → R, δF/δΠ = ∇F (Π). Thus, the (−) Lie–Poisson bracket is given via equation (9.4.19) by the triple product {F, G}(Π) = −Π · (∇F (Π) × ∇G(Π)).

(9.4.26)

Since δH/δΠ = ω, we see that for any F : R3 → R, d ˙ =Π ˙ · ∇F (Π) = −Π · (∇F (Π) × ω) (F (Π)) = DF (Π) · Π dt = ∇F (Π) · (Π × ω) so that F˙ = {F, H} for any F : R3 → R is equivalent to Euler’s equations of motion (9.4.24). The following result, due to Pauli [1953], Martin [1959], Arnol’d [1966], Sudarshan and Mukunda [1974] summarizes the situation. 9.4.6 Proposition. Euler’s equations (9.4.24) for a free rigid body are a Hamiltonian system in R3 relative to the (−) Lie Poisson bracket (9.4.24) and Hamiltonian function (9.4.25).  9.4.7 Example (Ideal Incompressible Homogeneous Fluid Flow). In §8.2 we have shown that the equations of motion for an ideal incompressible homogeneous fluid in a region Ω ⊂ R3 with smooth boundary ∂Ω are given by Euler’s equations of motion ∂v + (v · ∇)v = −∇p ∂t div v = 0 v(t, x) ∈ Tx (∂Ω) for x ∈ ∂Ω

(9.4.27a) (9.4.27b) (9.4.27c)

with initial condition v(0, x) = v0 (x), a given vector field on Ω. Here v(t, x) is the Eulerian or spatial velocity, a time dependent vector field on Ω. The pressure p is a function of v and is uniquely determined by v (up to a constant) by the Neumann problem (take div and the dot product with n of the first equation in (9.4.27)) ∆p = − div((v · ∇)v) ∂p = ∇p · n = −((v · ∇)v) · n on ∂Ω, ∂n where n is the outward unit normal to ∂Ω. The kinetic energy Z 1 H(v) = kvk2 d3 x 2 Ω

(9.4.28a) (9.4.28b)

(9.4.29)

574

9. Applications

has been shown in §8.2 to be a conserved quantity for (9.4.27). We shall prove below that the first equation in (9.4.27) is Hamiltonian relative to a (+) Lie–Poisson bracket with Hamiltonian function given by (9.4.29). Consider the Lie algebra Xdiv (Ω) of divergence free vector fields on Ω tangent to ∂Ω with bracket given by minus the bracket of vector fields, that is, for u, v ∈ Xdiv (Ω) define [u, v] = (v · ∇)u − (u · ∇)v.

(9.4.30)

(The reason for this strange choice comes from the fact that the usual Lie bracket for vector fields is the right Lie algebra bracket of the diffeomorphism group of Ω.) Now pair Xdiv (Ω) with itself via the L2 -pairing. As in Example 9.4.2, using the Hodge–Helmholtz decomposition, it follows that this pairing is weakly nondegenerate. In particular δH = v. (9.4.31) δv The (+) Lie–Poisson bracket on Xdiv (Ω) is      Z δF δF δG 3 δG ·∇ − ·∇ d x. (9.4.32) {F, G}(v) = v· δv δv δv δv Ω Therefore, for any Xdiv (Ω) → R, we have Z d δF ˙ 3x (F (v)) = DF (v) · v˙ = · vd dt Ω δv     Z δF δF = v · (v · ∇) − · ∇ v d3 x δv δv Ω     Z Z ∂F 1 δF 3 2 d x− ·∇ kvk d3 x. = v · (v · ∇) δv 2 Ω ∂v Ω To handle the first integral, observe that if f, g : Ω → R

div(f gv) = f div(gv) + gv · ∇f,

then

so that by Stokes’ theorem and v · n = 0, div v = 0, we get Z Z Z gv · ∇f d3 x = f g v · n dS − f div(g v)d3 x Ω ∂Ω Ω Z =− f v · ∇g d3 x. Ω

Applying the above relation to g = v i , f = δF/δvi , and summing over i = 1, 2, 3 we get   Z Z δF δF 3 v · (v · ∇) d x=− · ((v · ∇)v)d3 x δv Ω Ω δv so that F˙ = {F, H} reads Z Ω

δF · v˙ d3 x = − δv

Z Ω

  δF 1 · (v · ∇)v + ∇kvk2 d3 x δv 2

(9.4.33)

for any F : Xdiv (Ω) → R. One would like to conclude from here that the coefficients of δF/δv on both sides of equation (9.4.33) are equal. This conclusion, however, is incorrect, since 1 (v · ∇)v + ∇kvk2 2

9.4 The Lie–Poisson Bracket in Continuum Mechanics and Plasmas

575

is not divergence free. Thus, applying the Hodge–Helmholtz decomposition, write 1 (v · ∇)v + ∇kvk2 = X − ∇f 2

(9.4.34)

where X ∈ Xdiv (Ω) and f is determined by   1 ∆ f + kvk2 = − div((v · ∇)v), and 2   ∂ 1 f + kvk2 = −((v · ∇) · v) · n ∂n 2 which coincides with equation (9.4.28), that is, 1 f + kvk2 = p + constant. 2

(9.4.35)

Moreover, since Z Ω

δF · ∇f d3 x = δv

Z

∂F f · n dS − ∂Ω ∂v

Z f div Ω

δF 3 d x=0 δv

we have from equations (9.4.33), (9.4.34), (9.4.35) ˙ ∂v = −X = −(v · ∇)v = ∇p ∂t which is the first equation in equation (9.4.27). We have thus proved the following result (see Arnol’d [1966], Marsden and Weinstein [1983]). 9.4.8 Proposition. Euler’s equations (9.4.27) are a Hamiltonian system on Xdiv (Ω) relative to the (+) Lie–Poisson bracket (9.4.32) and Hamiltonian function given by (9.4.29).  9.4.9 Example (The Poisson–Vlasov Equation). We consider a collisionless plasma consisting (for notational simplicity) of only one species of particles with charge q and mass m moving in Euclidean space R3 with positions x and velocities v. Let f (x, v, t) be the plasma density in the plasma space at time t. In the Coulomb or electrostatic case in which there is no magnetic field, the motion of the plasma is described by the Poisson–Vlasov equations which are the (collisionless) Boltzmann equations for the density function f and the Poisson equation for the scalar potential ϕf : ∂f ∂f q ∂ϕf ∂f +v· − · =0 ∂t ∂xZ m ∂x ∂v ∆ϕf = −q f (x, v)d3 v = ρf ,

(9.4.36) (9.4.37)

where ∂/∂x and ∂/∂v denote the gradients in R3 relative to the x and v variables, ρf is the charge density in physical space, and ∆ is the Laplacian. Equation (9.4.36) can be written in “Hamiltonian” form ∂f + {f, H} = 0, ∂t where { , } is the canonical Poisson bracket on phase space, namely,   ∂f ∂g ∂f ∂f 1 ∂f ∂g ∂f ∂g {f, g} = · − · = · − · , ∂x ∂p ∂p ∂x m ∂x ∂v ∂v ∂x

(9.4.38)

(9.4.39)

576

9. Applications

where p = mv and H = Hf = mkvk2 + qϕf is the single particle energy, called the self–consistent Hamiltonian. Indeed,     1 ∂Hf ∂f ∂Hf ∂f 1 ∂ϕf ∂f ∂f {Hf , f } = · − · = q · − mv · m ∂x ∂v ∂v ∂x m ∂x ∂v ∂x q ∂ϕf ∂f ∂f ∂f = · −v· = m ∂x ∂v ∂x ∂t according to equation (9.4.36). There is another very useful way to think of the evolution of f . If F (f ) is any functional of the density function f and f evolves according to the Poisson–Vlasov equations (9.4.36) (or (9.4.38) equivalently) then F evolves in time by F˙ = {F, H}+ where { , }+ is a (+) Lie–Poisson bracket (to be defined) of functionals and H is the total energy. Let us state this more precisely. Let  V = f ∈ C k (R6 ) | f → 0 as kxk → ∞, kvk → ∞ with the L2 -pairing h , i : V × V → R; Z hf, gi =

f (x, v)g(x, v)d3 xd3 v.

If F : V → R is differentiable at f ∈ V , the functional δF/δf is, by definition, the unique element δF/δf ∈ V such that  Z  δF δF ,g = (x, v)g(x, v) d3 x d3 v. DF (f ) · g = δf δf The vector space V is a Lie algebra relative to the canonical Poisson bracket (9.4.39) on R6 . For two functionals F, G : V → R their (+) Lie–Poisson bracket {F, G}+ : V → R is then given by   Z δF δG {F, G}+ (x, v) = f (x, v) , (x, v) d3 x d3 v, δf δf where { , } is the canonical bracket (9.4.39). For any f, g, h ∈ V we have the formula Z Z 3 3 f {g, h} d x d v = g{h, f } d3 x d3 v.

(9.4.40)

Indeed by integration by parts, we get Z Z Z ∂g ∂h 3 3 1 ∂h ∂g 3 3 1 f · d xd v − f · d xd v f {g, h} d3 x d3 v = m ∂x ∂v m ∂x ∂v Z Z 1 ∂f ∂h 3 3 1 ∂f ∂h 3 3 =− ·g d xd v + g · gd xd v m ∂x ∂v m ∂v ∂x  Z  1 ∂h ∂f ∂f ∂h = g · − · d3 x d3 v m ∂x ∂v ∂x ∂v Z = g{h, f } d3 x d3 v. We have the following results of Iw´ınski and Turski [1976], Kaufman and Morrison [1982] and Morrison [1980].

9.4 The Lie–Poisson Bracket in Continuum Mechanics and Plasmas

577

9.4.10 Proposition. Densities f ∈ V evolve according to the Poisson–Vlasov equation (9.4.38) if and only if any differentiable function F : V → R having functional derivative δF/δf evolves by the (+) Lie–Poisson equation F˙ (f ) = {F, H}+ (f ) (9.4.41) with the Hamiltonian H : V → R equal to the total energy 1 H(f ) = 2 Proof.

Z

2

3

3

Z

mkvk f (x, v) d x d v +

1 ϕf (x) d3 x. 2

First we compute δH/δf using the definition Z DH(f ) · δf =

δH δf. δf

Note that the first term of H(f ) is linear in f and the second term is Z Z 1 1 3 ϕ f ρf d x = k∇ϕf k2 d3 x 2 2 since ∆ϕf = −ρf . Using the chain rule and integration by parts, we get Z Z 1 3 3 2 DH(f ) · δf = mkvk δf d x d v + (∇ϕf )(D(∇ϕf ))δf d3 x 2 Z Z 1 = mkvk2 δf d3 x d3 v − ϕf D(∆ϕf ))δf d3 x 2   Z  Z Z 1 2 3 3 3 mkvk δf d x d v + ϕf D q f d v = (f )δf d3 x 2 Z Z 1 2 3 3 mkvk δf d x d v + ϕf qδf d3 v. = 2 Therefore, δH 1 = mkvk2 + qϕf = Hf . δf 2 We have F˙ (f ) = DF (f ) · f˙ =

Z

δH ˙ 3 3 f d xd v δf

and Z

  δF δH , f d3 x d 3 v δf δf  Z  δF δH , d3 x d3 v = {F, H}+ (f ). = f δf δf

δF {Hf , f } d3 x d3 v = δf

Z

by (9.4.40). Thus (9.4.41), for any F having functional derivatives, is equivalent to (9.4.38).

 

9.4.11 Example (The Maxwell–Vlasov Equations). We consider a plasma consisting of particles with charge q1 and mass m moving in Euclidean space R3 with positions x and velocities v. For simplicity we

578

9. Applications

consider only one species of particle; the general case is similar. Let f (x, v, t) be the plasma density at time t, E(x, t) and B(x, t) be the electric and magnetic fields. The Maxwell–Vlasov equations are:   ∂f ∂f q v×B ∂f +v· + E+ · = 0, (9.4.42a) ∂t ∂x m c ∂ v˙ 1 ∂B = − curl E, (9.4.42b) c ∂t Z 1 ∂E q vf (x, v, t) d3 v, = curl B − (9.4.42c) c ∂t c together with the non evolutionary equations Z div E = ρf , where ρf = q

f (x, v, t) d3 v,

div B = 0.

(9.4.43a) (9.4.43b)

Letting c → ∞ leads to the Poisson–Vlasov equation (9.4.36) ∂f q ∂ϕf ∂f ∂f +v· − · = 0, ∂t ∂x m ∂x ∂v where ∆ϕf = −ρf . In what follows we shall set q = m = c = 1. The Hamiltonian for the Maxwell–Vlasov system is Z Z 1 1 H(f, E, B) = kvk2 f (x, v, t) dx dv + [kE(x, t)k2 + kB(x, t)k2 ] d3 x. (9.4.44) 2 2 Let V = { curl Z | Z ∈ X(R3 ) }. We have the following result of Iw´ınski and Turski [1976], Morrison [1980], and Marsden and Weinstein [1982] 9.4.12 Theorem.

(i) The manifold F(R6 ) × X(R3 ) × V is a Poisson manifold relative to the bracket  Z  δF δG {F, G}(f, E, B) = f , d3 x d3 v δf δf  Z  δG δG δG δF + · curl − · curl d3 x δB δB δE δB  Z  δF ∂f δG δG ∂f δF + · − · d3 xd3 v δE ∂v δf δE ∂v δf   Z ∂ δF ∂ δG + fB · × d3 x d3 v. (9.4.45) ∂v δf ∂v δf

(ii) The equations of motion (9.4.42) are equivalent to F˙ = {F, H}

(9.4.46)

where F is any locally defined function with functional derivatives and { , } is given by 9.4.44. Proof. Part (i) follows from general considerations on reduction (see Marsden and Weinstein [1982]). The direct verification is laborious but straightforward, if one recognizes that the first two terms are the Poisson bracket for the Poisson–Vlasov equation and the Born–Infeld bracket respectively. (ii) Since δH 1 = kvk2 , δf 2

δH = E, δE

and

curl

δH = curl B, δB

9.4 The Lie–Poisson Bracket in Continuum Mechanics and Plasmas

579

we have, by equation (9.4.40) and integration by parts in the fourth integral,   Z  Z  δF 1 δF δF 2 3 3 {F, H} = f , kvk d x d v + · curl B − E · curl d3 x δf 2 δB δB  Z  ∂f δF δF ∂f 1 2 · kvk − E · d3 x d3 v + δE ∂v 2 ∂v δE    Z ∂ δF ∂ 1 2 + fB × kvk d3 x d3 v ∂v δf ∂v 2    Z 1 δf δF 2 kvk , f − F · − divv (v × f B) d3 x d3 v = δf 2 δv   Z Z Z  ∂f 1 δF δF 2 3 3 · curl B + kvk d v d x − E · curl d3 x + δE ∂v 2 δB where divv denotes the divergence only with respect to the v-variable. Since   1 ∂f ∂f 2 kvk , f = −v · , divv (v × f B) = · (v × B), 2 ∂x ∂v Z Z ∂f 1 kvk2 d3 x = − vf d3 x, ∂v 2 and

Z E · curl

δF 3 d x= δB

Z

δF · curl E d3 x, δB

we get   ∂f ∂f 3 3 δF −v · − (E + v × B) · d xd v {F, H} = δf ∂x ∂v   Z Z δF + · curl B − vf (x, v, t)d3 v d3 v δE  Z  δF − · curl E d3 x, δB Z

and since F˙ =

Z

δF ˙ 3 3 · f d xd v + δf

Z

δF ˙ 3 · Ed x + δE

Z

(9.4.47)

δ0 F ˙ 3 · Bd x δB

taking into account that δF/δB, δ 0 F/δB differ by a gradient (by equation (9.4.15)) which is L2 -orthogonal to V (of which both curl E and B are a member), it follows from equation (9.4.47) that the equations (9.4.42)–(9.4.43) (with q = c = m = 1) are equivalent to (9.4.46).  

Exercises  9.4-1.

Find the symplectic form equivalent to the Born–Infeld bracket (9.4.16) on V × X(R3 ).

 9.4-2. Show that the Hamiltonian vector field XH ∈ X(g∗ ) relative to the (±) Lie–Poisson bracket is given by XH (µ) = ∓ ad(δH/δµ)∗ µ.  9.4-3 (Gardner [1971]).

Let V = { f : R → R | f is C ∞ , lim|x|→∞ f (x) = 0 }.

580

9. Applications

(i) Show that the prescription Z

+∞

{F, G}(f ) = −∞

δF d δG dx δf dx δf

defines a Poisson bracket on V for appropriate functions F and G (be careful about what hypotheses you put on F and G). (ii) Show that the Hamiltonian vector field of H : V → R is given by XH (f ) =

d δH . dx δf

R +∞ (iii) Let H(f 0 ) = −∞ (f 3 + (1/2)fx2 )dx. Show that the differential equation for XH is the Korteweg– deVries equation: ft − 6f fx + fxxx = 0.  9.4-4.

Let g be a Lie algebra and  ∈ g∗ be fixed. Show that the prescription    δF δG , {F, G} (µ) = , δµ δµ

defines a Poisson bracket on g∗ . Hint: Look at the formulas in the proof of Theorem 9.4.4.  9.4-5. (i) (Pauli [1953], Jost [1964]). Let P be a finite dimensional Poisson manifold satisfying the following condition: If f {F, G} = 0 for any locally defined F implies G = constant. Show that there exists an open dense set U in P such that the Poisson bracket restricted to U comes from a symplectic form on U . Hint: Define B : T ∗ P × T ∗ P → R by B(dF, dG) = {F, G}. Show first that U = { p ∈ P | Bp (α, β) = 0 for all α ∈ Tp∗ P implies βp = 0 } is open and dense in P . Then show that B can be inverted at points in U . (ii) Show that, in general, U 6= P by the following example. On R2 define  {F, G}(x, y) = y

∂F ∂G ∂F ∂G − ∂x ∂y ∂y ∂x



Show that U in (i) is R2 \{Ox-axis}. Show that on U , the symplectic form generating the above Poisson bracket is dx ∧ dy/y.

9.5

Constraints and Control

The applications in this final section all involve the Frobenius theorem. Each example is necessarily treated briefly, but hopefully in enough detail so the interested reader can pursue the subject further by utilizing the given references.

9.5 Constraints and Control

581

Constraints. We start with the subject of holonomic constraints in Hamiltonian systems. A Hamiltonian system as discussed in §8.1 can have a condition imposed that limits the available points in phase space. Such a condition is a constraint. For example, a ball tethered to a string of unit length in R3 may be considered to be constrained only to move on the unit sphere S 2 (or possibly interior to the sphere if the string is collapsible). If the phase space is T ∗ Q and the constraints are all derivable from constraints imposed only on the configuration space (the q’s), the constraints are called holonomic. For example, if there is one constraint f (q) = 0 for f : Q → R, the constraints on T ∗ Q can be simply obtained by differentiation: df = 0 on T ∗ Q. If the phase space is T Q, then the constraints are holonomic iff the constraints on the velocities are saying that the velocities are tangent to some constraint manifold of the positions. A constraint then can be thought of in terms of velocities as a subset E ⊂ T M . If it is a subbundle, this constraint is thus holonomic iff it is integrable in the sense of Frobenius’ theorem. Constraints that are not holonomic, are naturally called nonholonomic constraints. Holonomic constraints can be dealt with in the sense that one understands how to modify the equations of motion when the constraints are imposed, by adding forces of constraint, such as centrifugal force. See, for example Goldstein [1980, Chapter 1], and Abraham and Marsden [1978, Section 3.7]. We shall limit ourselves to the discussion of two examples of nonholonomic constraints. See Bloch, Ballieul, Crouch, and Marsden [2003] for an extensive discussion and background. A classical example of a nonholonomic system is a disk rolling without slipping on a plane. The disk of radius a is constrained to move without slipping on the (x, y)-plane. Let us fix a point P on the disk and call θ the angle between the radius at P and the contact point Q of the disk with the plane, as in Figure 9.5.1. Let (x, y, a) denote the coordinates of the center of the disk. Finally, if θ denotes the angle between the tangent line to the disk at Q and the x-axis, the position of the disk in space is completely determined by (x, y, θ, ϕ). These variables form elements of our configuration space M = R2 × S 1 × S 1 . The condition that there is no slipping at Q means that the velocity at Q is zero; that is, dθ dx + a cos ϕ = 0, dt dt

dy dθ + a sin ϕ = 0 dt dt

(total velocity = velocity of center plus the velocity due to rotation by angular velocity dθ/dt).

z

P y

Q

x

Figure 9.5.1. A rolling disk

These constraints may be written in terms of differential forms as ω1 = 0, ω2 = 0, where ω1 = dx + a cos ϕdθ

and ω2 = dy + a sin ϕ dθ.

582

9. Applications

We compute that ω = ω1 ∧ ω2 = dx ∧ dy + a cos ϕ dθ ∧ dy + a sin ϕ dx ∧ dθ, dω1 = −a sin ϕ dϕ ∧ dθ, dω2 = a cos ϕ dϕ ∧ dθ, dω1 ∧ ω = −a sin ϕ dϕ ∧ dθ ∧ dx ∧ dy, dω2 ∧ ω = a cos ϕ dϕ ∧ dθ ∧ dx ∧ dy. These do not vanish identically. Thus, according to Corollary 7.4.20, this system is not integrable and hence these constraints are nonholonomic. A second example of constraints is due to Nelson [1967]. Consider the motion of a car and denote by (x, y) the coordinates of the center of the front axle, ϕ the angle formed by the moving direction of the car with the horizontal, and θ the angle formed by the front wheels with the car (Figure 9.5.2).

y

x

Figure 9.5.2. Automobile maneuvers

The configuration space of the car is R2 × T2 , parameterized by (x, y, ϕ, θ). We shall prove that the constraints imposed on this motion are nonholonomic. Call the vector field X = ∂/∂θ steer . We want to compute a vector field Y corresponding to drive. Let the car be at the configuration point (x, y, ϕ, θ) and assume that it moves a small distance h in the direction of the front wheels. Notice that the car moves forward and simultaneously turns. Then the next configuration is (x + h cos(ϕ + θ) + o(h), y + h sin(ϕ + θ) + o(h), ϕ + h sin θ + o(h), θ). Thus the “drive” vector field is Y = cos(ϕ + θ)

∂ ∂ ∂ + sin(ϕ + θ) + sin θ . ∂x ∂y ∂ϕ

A direct computation shows that the vector field wriggle, W = [X, Y ] = − sin(ϕ + θ)

∂ ∂ ∂ + cos(ϕ + θ) + cos θ , ∂x ∂y ∂ϕ

and slide, S = [W, Y ] = − sin ϕ

∂ ∂ + cos ϕ , ∂x ∂y

satisfy [X, W ] = −Y,

[S, X] = 0,

[S, Y ] = sin θ cos ϕ

∂ ∂ + sin θ sin ϕ , ∂x ∂y

9.5 Constraints and Control

and [S, W ] = cos θ cos ϕ

583

∂ ∂ + cos θ sin ϕ . ∂x ∂y

Define the vector fields Z1 and Z2 by Z1 = [S, Y ] = −W + (cos θ)S + cos θ Z2 = [S, W ] = Y − (sin θ)S − sin θ

∂ , ∂ϕ

∂ . ∂ϕ

A straightforward calculation shows that [X, Z1 ] = Z2 ,

[X, Z2 ] − Z1 ,

[S, Z1 ] = 0,

[S, Z2 ] = 0,

[Z1 , Z2 ] = 0,

that is, {X, Z1 , Z2 , S} span a four dimensional Lie algebra g with one dimensional center spanned by S. In addition, its derived Lie algebra [g, g] ⊂ g, equals span {Z1 , Z2 } and is therefore abelian and two dimensional. Thus g has no nontrivial non-abelian Lie subalgebras. In particular the subbundle of T (R2 × T2 ) spanned by X and Y is not involutive and thus not integrable. By the Frobenius theorem, the field of two-dimensional planes spanned by X and Y is not tangent to a family of two-dimensional integral surfaces. Thus the motion of the car, subjected only to the constraints of “steer” and “drive” is nonholonomic. On the other hand, the motion of the car subjected to the constraints of “steer”, “drive” and “wriggle” is holonomic. Moreover, since the Lie algebra generated by these three vector fields is abelian, the motion of the car with these constraints can be described by applying these three vector fields in any order. Control. Next we turn our attention to some elementary aspects of control theory. We shall restrict our attention to a simple version of a local controllability theorem. For extensions and many additional results, we recommend consulting the book of and a few of the important papers and notes such as Brockett [1970, 1983], Sussmann [1977], Hermann and Krener [1977], Russell [1979], Hermann [1980], and Ball, Marsden, and Slemrod [1982] and references therein. Consider a system of differential equations of the form w(t) ˙ = X(w(t)) + p(t)Y (w(t))

(9.5.1)

on a time interval [0, T ] with initial conditions w(0) = w0 where w takes values in a Banach manifold M , X and Y are smooth vector fields on M and p : [0, T ] → R is a prescribed function called a control. The existence theory for differential equations guarantees that equation (9.5.1) has a flow that depends smoothly on w0 and on p lying in a suitable Banach space Z of maps of [0, T ] to R, such as the space of C 1 maps. Let the flow of (9.5.1) be denoted Ft (w0 , p) = w(t, p, w0 ).

(9.5.2)

We consider the curve w(t, 0, w0 ) = w0 (t); that is, an integral curve of the vector field X. We say that (9.5.1) is locally controllable (at time T ) if there is a neighborhood U of w0 (T ) such that for any point h ∈ U , there is a p ∈ Z such that w(T ; p, w0 ) = h. In other words, we can alter the endpoint of w0 (t) in a locally arbitrary way by altering p (see Figure 9.5). To obtain a condition under which local controllability can be guaranteed, we fix T and w0 and consider the map P : Z → M ; p 7→ w(T, p, w0 ). (9.5.3) The strategy is to apply the inverse function theorem to P . The derivative of Ft (w0 , p) with respect to p in the direction ρ ∈ Z is denoted Dp Ft (w0 , 0) · ρ = Lt ρ ∈ TFt (w0 ,0) M.

584

9. Applications

U h

Figure 9.5.3. Controllability

Differentiating d w(t, p) = X(w(t, p)) + p(t)Y (w(t, p)) dt with respect to p at p = 0, we find that in T 2 M d (9.5.4) Lt ρ = X(w0 (t)) · Lt ρ + (ρY (w0 (t)))vertical lift . dt To simplify matters, let us assume M = E is a Banach space and that X is a linear operator, so equation (9.5.4) becomes d Lt ρ = X · Lt ρ + ρY (w0 (t)). (9.5.5) dt Equation (9.5.5) has the following solution given by the variation of constants formula Z T LT ρ = e(T −s)X ρ(s)Y (esX w0 )ds (9.5.6) 0

since w0 (t) = etX w0 for linear equations. 9.5.1 Proposition. If the linear map LT : Z → E given by equation (9.5.6) is surjective, then equation (9.5.1) is locally controllable (at time T ). Proof. This follows from the “local onto” form of the implicit function theorem (see Theorem 2.5.9) applied to the map P . Solutions exist for time T for small p since they do for p = 0; see Corollary 4.1.25.  9.5.2 Corollary.

Suppose E = Rn and Y is linear as well. If dim span {Y (w0 ), [X, Y ](w0 ), [X, [X, Y ]](w0 ), . . . } = n,

then equation (9.5.1) is locally controllable. Proof.

We have the Baker–Campbell–Hausdorff formula e−sX Y esX = Y + s[X, Y ] +

s2 [X, [X, Y ]] + · · · , 2

obtained by expanding esX = I + sX + (s2 /2)X 2 + · · · and gathering terms. Substitution into equation (9.5.6) shows that LT is surjective.  For the case of nonlinear vector fields and the system equation (9.5.1) on finite-dimensional manifolds, controllability hinges on the dimension of the space obtained by replacing the foregoing commutator brackets by Lie brackets of vector fields, n being the dimension of M . This is related to what are usually called Chow’s theorem in control theory (see Chow [1947]). To see that some condition involving brackets is necessary, suppose that the span of X and Y forms an involutive distribution of T M . Then by the Frobenius theorem, w0 lies in a unique maximal two-dimensional leaf £(w0 ) of the corresponding foliation. But then the solution of equation (9.5.1) can never leave £(w0 ), no matter how p is chosen. Hence in such a situation, equation (9.5.1) would not be locally controllable; rather, one would only be able to move in a two-dimensional subspace. If repeated bracketing with X increases the dimension of vectors obtained then the attainable states increase in dimension accordingly.

9.5 Constraints and Control

585

Exercises  9.5-1. Check that the system in Figure 9.5.1 is nonholonomic by verifying that there are two vector fields X, Y on M spanning the subset E of T M defined by the constraints x˙ = a θ˙ cos ϕ = 0 and y˙ + a θ˙ sin ϕ = 0 such that [X, Y ] is not in E; that is, use Frobenius’ theorem directly rather than using Pfaffian systems.  9.5-2. Justify the names wriggle and slide for the vector fields W and S in the example of Figure 9.5.2 using the product formula in Exercise 4.2-5. Use these formulas to explain the following statement of Nelson [1967, p. 35]: “the Lie product of “steer” and “drive” is equal to “slide” and “rotate” (= ∂/∂ϕ) on θ = 0 and generates a flow which is the simultaneous action of sliding and rotating. This motion is just what is needed to get out of a tight parking spot.”  9.5-3. The word holonomy arises not only in mechanical constraints as explained in this section but also in the theory of connections (Kobayashi and Nomizu [1963, Volume II, Sections 7 and 8]). What is the relation between the two uses, if any?  9.5-4.

In linear control theory equation (9.5.1) is replaced by w(t) ˙ = X · w(t) +

N X

pi (t)Yi ,

i=1

where X is a linear vector field on Rn and Yi are constant vectors. By using the methods used to prove Proposition 9.5.1, rediscover for yourself the Kalman criterion for local controllability, namely, the set  k X Yi | k = 0, 1, . . . , n − 1, i = 1, . . . , N spans Rn .

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